Similarity to a number of online. Numerical rows: appointment, authority, signs of wealth, applications, decisions 1 3n n 3 2
Tsya stattya є structured and reported information, as it is possible in a good time for the analysis of the rights and the task. Let's take a look at the topic of number series.
Tsya article begins with the main functions to understand. We gave standard options and vivimo basic formulas. In order to close the material, the main application has been put in the article.
Basic theses
We can represent the system: a 1 , a 2 . . . , a n , . . . de a k ∈ R , k = 1 , 2 . . . .
For example, take the following numbers, like: 6, 3, - 3 2, 3 4, 3 8, - 3 16, . . . .
Appointment 1
The number series is the sum of terms ∑ ak k = 1 ∞ = a 1 + a 2 + . . . + a n +. . . .
In order to better understand the meaning, we can look at the vipadok, for which q \u003d - 0. 5: 8 - 4 + 2 - 1 + 1 2 - 1 4 +. . . = ∑ k = 1 ∞ (- 16) · - 1 2 k .
Appointment 2
a k k-im low member.
Vіn looking like this rank - 16 · - 1 2 k.
Appointment 3
Chastkov's sum in a row looks like this order Sn = a1+a2+. . . + a n , yakіy n-Be it a number. S n nth the sum is low.
For example, ∑ k = 1 ∞ (- 16) · - 1 2 k є S 4 = 8 - 4 + 2 - 1 = 5 .
S 1 , S 2 , . . . , S n , . . . utvoryuyuyut inconsistency sequence of the numerical series.
For a row n-a the sum is behind the formula S n = a 1 (1 - q n) 1 - q = 8 1 - - 1 2 n 1 - - 1 2 = 16 3 1 - - 1 2 n. Victoriously, the sequence of private sums will come: 8, 4, 6, 5, . . . , 16 3 1 - - 1 2 n , . . . .
Appointment 4
Series ∑ k = 1 ∞ a k є similar then, if the sequence can be the end of the line S = lim S n n → + ∞ . If there is no boundary, or the sequence is not limited, then the series ∑ k = 1 ∞ a k is called rozbіzhnym.
Appointment 5
Sumy row, what to go∑ k = 1 ∞ a k
For this application lim S n n → + ∞ = lim 16 3 t → + ∞ 1 - 1 2 n = 16 3 lim n → + ∞ 1 - - 1 2 n = 16 3 , row ∑ k = 1 ∞ (- 16) · - 1 2 k converge. The sum is expensive 16 3: ∑ k = 1 ∞ (- 16) · - 1 2 k = 16 3 .
butt 1
As a butt of a row, you can draw a sum of geometric progression with a larger banner, lower one: 1 + 2 + 4 + 8 +. . . + 2n - 1 +. . . = ∑ k = 1 ∞ 2 k - 1 .
n-a part sum is determined by the virase S n = a 1 (1 - q n) 1 - q = 1 (1 - 2 n) 1 - 2 = 2 n - 1, and the inter-part sum is not limited: lim n → + ∞ S n = lim n → + ∞ (2 n - 1) = + ∞ .
Another example of a random number series is the sum of the form ∑ k = 1 ∞ 5 = 5 + 5 + . . . . For this account, n-a part of the sum can be calculated as Sn = 5n. The inter-partial sums are not limited lim n → + ∞ S n = lim n → + ∞ 5 n = + ∞.
Appointment 6
The sum of this form is yak ∑ k = 1 ∞ = 1 + 1 2 + 1 3 + . . . + 1n +. . . – ce harmonious number row.
Appointment 7
Sum ∑ k = 1 ∞ 1 k s = 1 + 1 2 s + 1 3 s +. . . + 1ns + . . . , de s –deisne number, є zagalnenno harmonic number series.
Appointments, looked at more, will help you to compose more applications and orders.
In order to complete the appointment, it is necessary to bring the line equal.
- ∑ k = 1 ∞ 1 k
Diemo by the method of reversal. If the wines converge, then the border is skinny. You can write equal as lim n → + ∞ S n = S and lim n → + ∞ S 2 n = S . After singing diy we are obsessed with equality l i m n → + ∞ (S 2 n - S n) = 0 .
Navpaki,
S 2 n - S n \u003d 1 + 1 2 + 1 3 +. . . + 1n + 1n + 1 + 1n + 2 +. . . + 1 2 n - - 1 + 1 2 + 1 3 +. . . + 1n = 1n + 1 + 1n + 2 +. . . + 1 2 n
The just-so inconsistencies are 1 n + 1 > 1 2 n , 1 n + 1 > 1 2 n . . . , 1 2 n - 1 > 1 2 n . Come out, S 2 n - S n = 1 n + 1 + 1 n + 2 + . . . + 1 2 n > 1 2 n + 1 2 n +. . . + 1 2 n = n 2 n = 1 2 . Viraz S 2 n - S n > 1 2 to say that lim n → + ∞ (S 2 n - S n) = 0 is out of reach. A number of rozbіzhny.
- b1+b1q+b1q2+. . . + b 1 q n +. . . = ∑ k = 1 ∞ b 1 q k - 1
It is necessary to confirm that the sum of the sequence of numbers goes off at q< 1 , и расходится при q ≥ 1 .
Zgіdno with the help of the appointed persons, the sum n members are dependent on the formula S n = b 1 · (q n - 1) q - 1 .
Yakscho q< 1 верно
lim n → + ∞ S n = lim n → + ∞ b 1 q n - 1 q - 1 = b 1 lim n → + ∞ q n q - 1 - lim n → + ∞ 1 q - 1 = = b 1 0 - 1 q - 1 = b 1 q - 1
We brought that the number series converged.
For q = 1 b 1 + b 1 + b 1 +. . . ∑ k = 1 ∞ b 1 . Sumi can be known from the additional formula S n = b 1 · n , inter-infinite lim n → + ∞ S n = lim n → + ∞ b 1 · n = ∞ . In this variant, the row diverge.
Yakscho q = - 1 the row looks like b 1 - b 1 + b 1 - . . . = ∑ k = 1 ∞ b 1 (-1) k + 1 . Often sums look like S n = b 1 for unpaired n, i S n = 0 for guys n. Having looked at this vipadok, we are reconsidered that there are no gaps and a number of differences.
For q > 1, lim n → + ∞ S n = lim n → + ∞ b 1 (q n - 1) q - 1 = b 1 lim n → + ∞ q n q - 1 - lim n → + ∞ 1 q - 1 = = b 1 ∞ - 1 q - 1 = ∞
Mi brought, scho number series to diverge.
- The series ∑ k = 1 ∞ 1 k s converge so that s > 1 and diverge, so that s ≤ 1 .
For s = 1 we take ∑ k = 1 ∞ 1 k , the series diverge.
For s< 1 получаем 1 k s ≥ 1 k для k ,natural number. Oskіlki row є razbіzhnym ∑ k = 1 ∞ 1 k , then there is no difference. In addition to this, the sequence ∑ k = 1 ∞ 1 k s is uncircumscribed. Robimo wisnovok s< 1 .
It is necessary to prove that the series ∑ k = 1 ∞ 1 k s converges when s > 1.
Imagine S 2 n - 1 - S n - 1:
S 2 n - 1 - S n - 1 \u003d 1 + 1 2 s + 1 3 s +. . . + 1 (n - 1) s + 1 ns + 1 (n + 1) s + . . . + 1 (2 n - 1) s - - 1 + 1 2 s + 1 3 s +. . . + 1 (n - 1) s = 1 ns + 1 (n + 1) s + . . . + 1(2n - 1)s
Assume that 1 (n + 1) s< 1 n s , 1 (n + 2) s < 1 n s , . . . , 1 (2 n - 1) s < 1 n s , тогда S 2 n - 1 - S n - 1 = 1 n s + 1 (n + 1) s + . . . + 1 (2 n - 1) s < < 1 n s + 1 n s + . . . + 1 n s = n n s = 1 n s - 1
Representable equality for numbers that are natural and equal n = 2: S 2 n - 1 - S n - 1 = S 3 - S 1 = 1 2 s + 1 3 s< 1 2 s - 1 n = 4: S 2 n - 1 - S n - 1 = S 7 - S 3 = 1 4 s + 1 5 s + 1 6 s + 1 7 s < 1 4 s - 1 = 1 2 s - 1 2 n = 8: S 2 n - 1 - S n - 1 = S 15 - S 7 = 1 8 s + 1 9 s + . . . + 1 15 s < 1 8 s - 1 = 1 2 s - 1 3 . . .
We take:
∑ k = 1 ∞ 1 k s = 1 + 1 2 s + 1 3 s + 1 4 s +. . . + 1 7 s + 1 8 s +. . . + 1 15 s +. . . \u003d \u003d 1 + S 3 - S 1 + S 7 - S 3 + S 15 + S 7 +. . .< < 1 + 1 2 s - 1 + 1 2 s - 1 2 + 1 2 s - 1 3 + . . .
Viraz 1 + 1 2 s - 1 + 1 2 s - 1 2 + 1 2 s - 1 3 +. . . - The sum of the geometric progress q = 1 2 s - 1 . Zgіdno with vihіdnimi dannym at s > 1, then 0< q < 1 . Получаем, ∑ k = 1 ∞ < 1 + 1 2 s - 1 + 1 2 s - 1 2 + 1 2 s - 1 3 + . . . = 1 1 - q = 1 1 - 1 2 s - 1 . Последовательность ряда при s > 1 zbіlshuєtsya and intermingles with the beast 11-12s-1. Obviously, є between and a row є ∑ k = 1 ∞ 1 k s .
Appointment 8
Series ∑ k = 1 ∞ a k positive for that guy, so that the term > 0 ak > 0 , k = 1 , 2 , . . . .
Series ∑ k = 1 ∞ b k the sign is drawn as if the signs of numbers are vіdrіznyayutsya. Danish application of representations yak ∑ k = 1 ∞ b k = ∑ k = 1 ∞ (-1) k a k or ∑ k = 1 ∞ b k = ∑ k = 1 ∞ (- 1) k + 1 a k , de a k > 0 , k = 1, 2,. . . .
Series ∑ k = 1 ∞ b k familiar, to that in a new number of numbers, negative and positive.
The other option is a row - the last line of the third option.
Let's put it on for skin retraction:
6 + 3 + 3 2 + 3 4 + 3 8 + 3 16 + . . . 6 - 3 + 3 2 - 3 4 + 3 8 - 3 16 + . . . 6 + 3 - 3 2 + 3 4 + 3 8 - 3 16 + . . .
For the third option, it is also possible to designate the absolute mental comfort.
Appointment 9
The black sign of the series ∑ k = 1 ∞ b k is absolutely correct in that case, if ∑ k = 1 ∞ b k is also considered similar.
Reportedly, we analyze the sprat of characteristic options
butt 2
Yakscho row 6 - 3 + 3 2 - 3 4 + 3 8 - 3 16 +. . . i 6 + 3 - 3 2 + 3 4 + 3 8 - 3 16 +. . . appear as similar, then correctly enter that 6 + 3 + 3 2 + 3 4 + 3 8 + 3 16 +. . .
Appointment 10
The familiar series ∑ k = 1 ∞ b k is considered to be mentally similar to that one, since ∑ k = 1 ∞ b k is different, and the series ∑ k = 1 ∞ b k is considered similar.
butt 3
We report on the option ∑ k = 1 ∞ (-1) k + 1 k = 1 - 1 2 + 1 3 - 1 4 + . . . . The series ∑ k = 1 ∞ (- 1) k + 1 k = ∑ k = 1 ∞ 1 k , which is composed of absolute values, is chosen as a variant. This option is important to go, so it's easy to figure out. From the first example, we know that the series ∑ k = 1 ∞ (-1) k + 1 k = 1 - 1 2 + 1 3 - 1 4 + . . . bude vvazhatisya mentally similar.
Features of the rows that converge
Let's analyze the power for singing moods
- If ∑ k = 1 ∞ a k will converge, then the i series ∑ k = m + 1 ∞ a k is also recognized as such that it descends. You can specify which row without m members are also considered similar. If we add, if we add to ∑ k = m + 1 ∞ a k a bunch of numbers, then the result will also be similar.
- How ∑ k = 1 ∞ a k converge i sum = S, then converge i series ∑ k = 1 ∞ A a k , ∑ k = 1 ∞ A a k = A S , de A- Stay.
- How ∑ k = 1 ∞ a k and ∑ k = 1 ∞ b k є similar, sumi Aі B tezh, those rows ∑ k = 1 ∞ a k + b k i ∑ k = 1 ∞ a k - b k also converge. Sumi dorivnyuvatimut A+Bі A-B obviously.
Determine which series to go off ∑ k = 1 ∞ 2 3 k · k 3 .
Let us change ∑ k = 1 ∞ 2 3 k · k 3 = ∑ k = 1 ∞ 2 3 · 1 k 4 3 . The row ∑ k = 1 ∞ 1 k 4 3 is considered similar, but the row ∑ k = 1 ∞ 1 k s go off at s > 1. Depending on the other power, ∑ k = 1 ∞ 2 3 · 1 k 4 3 .
butt 5
Let the series ∑ n = 1 ∞ 3 + n n 5 2 converge.
Reversible cob variant ∑ n = 1 ∞ 3 + n n 5 2 = ∑ n = 1 ∞ 3 n 5 2 + n n 2 = ∑ n = 1 ∞ 3 n 5 2 + ∑ n = 1 ∞.
We subtract the sum ∑ n = 1 ∞ 3 n 5 2 and ∑ n = 1 ∞ 1 n 2 . The leather series is recognized as such that it is possible to descend to authority. The shards of the row converge, then the exit option is the same.
butt 6
Calculate how the series 1 - 6 + 1 2 - 2 + 1 4 - 2 3 + 1 8 - 2 9 + converge. . . and calculate the amount.
Exit option:
1 - 6 + 1 2 - 2 + 1 4 - 2 3 + 1 8 - 2 9 +. . . == 1 + 1 2 + 1 4 + 1 8 +. . . - 2 3 + 1 + 1 3 + 1 9 +. . . = = ∑ k = 1 ∞ 1 2 k - 1 - 2 ∑ k = 1 ∞ 1 3 k - 2
The leather series converge, the shards are one of the members of the numerical sequence. Vіdpovіdno to the third dominion, we can count, scho vihіdny variant is also similar. The sum is calculated: The first term of the series is ∑ k = 1 ∞ 1 2 k - 1 = 1, and the standard = 0 . 5 , then follows, ∑ k = 1 ∞ 1 2 k - 1 = 1 1 - 0 . 5 = 2. The first term ∑ k = 1 ∞ 1 3 k - 2 = 3 , and the sign of the descending numerical sequence = 1 3 . We take: ∑ k = 1 ∞ 1 3 k - 2 = 3 1 - 1 3 = 9 2 .
Vykoristuєmo virazi, obsessed more, in order to calculate the sum 1 - 6 + 1 2 - 2 + 1 4 - 2 3 + 1 8 - 2 9 +. . . = ∑ k = 1 ∞ 1 2 k - 1 - 2 ∑ k = 1 ∞ 1 3 k - 2 = 2 - 2 9 2 = - 7
Necessary intelligence for the appointment, chi є a number of similar
Appointment 11If the series ∑ k = 1 ∞ ak є is similar, then k-th term = 0: lim k → + ∞ a k = 0 .
If we are to believe it, whether it is a variant, it is necessary not to forget about the non-authentic mind. If it doesn’t win, the row will disperse. Like lim k → + ∞ a k ≠ 0 , the series is different.
Next, specify what the mind is important, but it’s not enough. Since the equality lim k → + ∞ a k = 0 wins, it does not guarantee that ∑ k = 1 ∞ a k is similar.
Let's give an example. For the harmonic series ∑ k = 1 ∞ 1 k, the Umoff vikonuetsya lim k → + ∞ 1 k = 0 , but the series still diverge.
butt 7
Calculate the efficiency ∑ n = 1 ∞ n 2 1 + n .
Let's reconsider lim n → + ∞ n 2 1 + n = lim n → + ∞ n 2 n 2 1 n 2 + 1 n = lim n → + ∞ 1 1 n 2 + 1 n = 1 + 0 + 0 = + ∞ ≠ 0
Mezha nth member is not good 0 . Mi brought, scho tsey row to disperse.
How to designate the zbіzhnіst sign-positive series.
How to constantly rank with the assigned signs, to be able to constantly count the boundaries. Tsej razdіl added to help stow away folded pіd hour vypіshennya priklіv that zavdan. To designate the zbіzhnіst sign-positive row, іsnuє pevna umova.
For a positive sign ∑ k = 1 ∞ a k , ak > 0 ∀ k = 1 , 2 , 3 , . . . It is necessary to calculate the amount of sums.
Yak porivnyuvati ranks
Іsnuє kіlka is a sign of the alignment of rows. Mi porіvnyuєmo row, zbіzhnіst kakogo proponuetsya vznáchiti, іz tim near, zbіzhnіst yak vіdoma.
Persha sign
∑ k = 1 ∞ a k and ∑ k = 1 ∞ b k - positive sign series. Unevenness a k ≤ b k is valid for k = 1, 2, 3, ... We can take ∑ k = 1 ∞ a k in the series ∑ k = 1 ∞ b k . Oskіlki ∑ k = 1 ∞ a k diverge, the series ∑ k = 1 ∞ b k can be taken as a divergence.
This rule is constantly victorious for vyshennya rivnyan and є seroznym argument, which will help to signify zbіzhnist. Skladnoshchi can lie in the fact that you need to take a butt for a match, you can know far from being in a skin depression. To finish often a number is chosen according to the principle k-th a member of the dorіvnyuvatime to the result of vіdnіmannya pokaznіvіv stаіnіv stаіv nіdnik і znamennik k-th members are low. It is acceptable that a k \u003d k 2 + 3 4 k 2 + 5 2 – 3 = - 1 . At to this particular type you can determine which row is necessary for alignment k-im member b k = k - 1 = 1 k, which is harmonious.
To close the material, let's take a look at a couple of typical options in detail.
butt 8
Significantly, yakim is the series ∑ k = 1 ∞ 1 k - 1 2 .
Shards of boundary = 0 lim k → + ∞ 1 k - 1 2 = 0 mi viconals necessary mind. Unevenness will be fair 1 k< 1 k - 1 2 для k , yakі є natural. From the previous paragraphs, we recognized that the harmonic series ∑ k = 1 ∞ 1 k is different. With the first sign, it can be brought to light that the final option is rozbіzhnym.
butt 9
Significantly, chi є a row similar or different ∑ k = 1 ∞ 1 k 3 + 3 k - 1 .
For which butt there is a need for intelligence, shards lim k → + ∞ 1 k 3 + 3 k - 1 = 0 . Serve at the sight of unevenness 1 k 3 + 3 k - 1< 1 k 3 для любого значения k. The series ∑ k = 1 ∞ 1 k 3 is similar, but the harmonic series ∑ k = 1 ∞ 1 k s converge when s > 1. Zgidno with the first sign, we can create visnovok, that the number series is similar.
butt 10
Vznachiti, yakim є series ∑ k = 3 ∞ 1 k ln (ln k) . lim k → + ∞ 1 k ln (ln k) = 1 + ∞ + ∞ = 0 .
For whom, all options can be called vikonannya necessary mind. Significantly a number of differences. For example, ∑ k = 1 ∞ 1 k s . In order to determine why the foot is good, we can look at the sequence (ln (ln k)), k = 3, 4, 5. . . . Sequence members ln (ln 3), ln (ln 4), ln (ln 5),. . . zbіshuєtsya to infinity. After analyzing the equality, we can say that, taking the role of the value N = 1619, then the sequence members > 2. For this sequence, the inequality 1 k ln (ln k) will be valid< 1 k 2 . Ряд ∑ k = N ∞ 1 k 2 сходится согласно первому признаку, так как ряд ∑ k = 1 ∞ 1 k 2 тоже сходящийся. Отметим, что согласно первому признаку ряд ∑ k = N ∞ 1 k ln (ln k) сходящийся. Можно сделать вывод, что ряд ∑ k = 3 ∞ 1 k ln (ln k) также сходящийся.
Another badge
Assume that ∑ k = 1 ∞ a k and ∑ k = 1 ∞ b k are positive sign numerical series.
If lim k → + ∞ a k b k ≠ ∞ , then the series ∑ k = 1 ∞ b k converge, i ∑ k = 1 ∞ a k converge also.
If lim k → + ∞ a k b k ≠ 0 , if the series ∑ k = 1 ∞ b k diverge, then ∑ k = 1 ∞ ak also diverge.
If lim k → + ∞ a k b k ≠ ∞ i lim k → + ∞ a k b k ≠ 0 , then the scalability of the scaling of the series means the scaling of the scaling of the other.
Let's take a look at ∑ k = 1 ∞ 1 k 3 + 3 k - 1 for other signs. For alignment ∑ k = 1 ∞ b k take the series ∑ k = 1 ∞ 1 k 3 . Significantly between: lim k → + ∞ k b k = lim k → + ∞ 1 k 3 + 3 k - 1 1 k 3 = lim k → + ∞ k 3 k 3 + 3 k - 1 = 1
With another sign, one can indicate that the series ∑ k = 1 ∞ 1 k 3, which converges, means that the cob variant also converges.
butt 11
Significantly, yakim is the series ∑ n = 1 ∞ k 2 + 3 4 k 3 + 5 .
Let's analyze the necessary mind lim k → ∞ k 2 + 3 4 k 3 + 5 = 0 , as in this variant it is victorious. Similar to another sign, take the series ∑ k = 1 ∞ 1 k . Shukaєmo between: lim k → + ∞ k 2 + 3 4 k 3 + 5 1 k = lim k →
Zgіdno with guiding theses, a row that diverges, pulling itself apart in a row of exits.
third mark
Let's look at the third sign of a break.
Assume that ∑ k = 1 ∞ a k and _ ∑ k = 1 ∞ b k are positive sign numerical series. If it is wise to calculate for the next number a k + 1 a k ≤ b k + 1 b k , then the efficiency of this series ∑ k = 1 ∞ b k means that the series ∑ k = 1 ∞ ak is also similar. Razbіzhny row ∑ k = 1 ∞ a k drag behind him razbіzhnіst ∑ k = 1 ∞ b k .
Sign of d'Alembert
Assume that ∑ k = 1 ∞ a k is a positive-sign number series. How lim k → + ∞ a k + 1 a k< 1 , то ряд является сходящимся, если lim k → + ∞ a k + 1 a k >1 then let's break it down.
Respect 1
The sign of d'Alembert is fair to that attitude, as the border is not narrow.
If lim k → + ∞ a k + 1 a k = - ∞ , then the series є is similar, if lim k → ∞ ak + 1 ak = + ∞ , then we divide.
If lim k → + ∞ ak + 1 ak = 1 , then the d'Alembert sign is not helpful and it is necessary to carry out more research.
butt 12
Significantly, chi є a row similar or different ∑ k = 1 ∞ 2 k + 1 2 k behind the d'Alembert sign.
It is necessary to reconsider, what is needed to win the mind. Let's calculate the distance, speeding up the Lopital rule: lim k → + ∞ 2 k + 1 2 k = ∞ ∞ = lim k → + ∞ 2 k + 1 "2 k" = lim k → + ∞ 2 2 k ln 2 = 2 + ∞ log 2 = 0
We can talk about what minds win. Using the d'Alembert sign: lim k → + ∞ = lim k → + ∞ 2 (k + 1) + 1 2 k + 1 2 k + 1 2 k = 1 2 lim k → + ∞ 2 k + 3 2 k + 1 = 12< 1
Row є similar.
butt 13
Significantly, chi є a row of arbitrarily ∑ k = 1 ∞ k k k ! .
We use the d'Alembert sign to show the difference in the series: lim k → + ∞ a k + 1 a k = lim k → + ∞ (k + 1) k + 1 (k + 1) ! k k k ! = lim k → + ∞ (k + 1) k + 1 k! k k · (k + 1)! = lim k → + ∞ (k + 1) k + 1 k k (k + 1) = = lim k → + ∞ (k + 1) k k k = lim k → + ∞ k + 1 k k = lim k → + ∞ 1 + 1 k k = e > 1
Otzhe, a number of є razbіzhnim.
Radical sign of Kosh
It is possible that ∑ k = 1 ∞ a k is a non-positive series. How lim k → + ∞ a k k< 1 , то ряд является сходящимся, если lim k → + ∞ a k k >1 then let's break it down.
Respect 2
If lim k → + ∞ ak k k = 1 , then this sign does not give the desired information - it is necessary to carry out additional analysis.
Tsya sign can be buti vikoristan in butts, yakі easily vyznachiti. Vipadok will be characteristic only if a member of the numerical series - tse showing stately viraz.
To close the otriman information, let's look at a sample of characteristic examples.
butt 14
Significantly, chi is a positive series ∑ k = 1 ∞ 1 (2 k + 1) k on similar.
It takes a mind to be respected by the vikonan, shards lim k → + ∞ 1 (2 k + 1) k = 1 + ∞ + ∞ = 0 .
Looking at the sign, looking through the eye, we can assume lim k → + ∞ a k k = lim k → + ∞ 1 (2 k + 1) k k = lim k → + ∞ 1 2 k + 1 = 0< 1 . Tsey rowє similar.
butt 15
Chi similar number series ∑ k = 1 ∞ 1 3 k · 1 + 1 k k 2 .
Vikorist's sign, described in the preceding paragraph lim k → + ∞ 1 3 k 1 + 1 k k 2 k = 1 3 lim k → + ∞ 1 + 1 k k = e 3< 1 , следовательно, числовой ряд сходится.
Integral sign of Koshi
Assume that ∑ k = 1 ∞ ak є is a positive sign series. It is necessary to designate the function of a non-permanent argument y = f(x), What is running a n = f (n) . Yakscho y = f(x) greater than zero, do not break and change to [a; + ∞) , where a ≥ 1
Then at the top, yakscho unclassified integral∫ a + ∞ f (x) d x є similar, then the series that you can see also converges. If the wines are separated, then in the butt a number of those are also separated.
When reversing the changed function, you can review the material reviewed in the previous lessons.
butt 16
Take a look at the stock ∑ k = 2 ∞ 1 k · ln k for feasibility.
Mindfulness in a row is respected by the vikonan, scaling lim k → + ∞ 1 k · ln k = 1 + ∞ = 0 . Let's look at y = 1 x ln x. Won is greater than zero, does not interrupt and changes to [2; +∞). The first two paragraphs are pre-determined, and on the third next, there is a report. We know better: y "= 1 x ln x" = x ln x "x ln x 2 = ln x + x 1 x x ln x 2 = - ln x + 1 x ln x 2. Won less for zero on [ 2 ; + ∞) It is not necessary to bring the thesis about those that the function is decaying.
Well, the function y = 1 x · ln x shows signs of the principle that we have seen more. Speeding it up: ∫ 2 + ∞ d x x ln x = lm A → + ∞ ∫ 2 A d (ln x) ln x = lim A → + ∞ ln (ln x) 2 A = = lim A → + ∞ (ln ( ln A) - ln (ln 2)) = ln (ln (+∞)) - ln (ln 2) = + ∞
Vіdpovіdno until otrimanih results, vyhіdny butt diverge, shards of unsound integration є razbіzhnym.
butt 17
Extend the series ∑ k = 1 ∞ 1 (10 k - 9) (ln (5 k + 8)) 3 .
Oskіlki lim k → + ∞ 1 (10 k - 9) (ln (5 k + 8)) 3 = 1 + ∞ = 0, then Umov is respected by the vikonana.
Starting from k = 4 , virniy viraz 1 (10 k - 9) (ln (5 k + 8)) 3< 1 (5 k + 8) (ln (5 k + 8)) 3 .
If the series ∑ k = 4 ∞ 1 (5 k + 8) (ln (5 k + 8)) 3 will be considered similar, then, according to one of the principles of alignment, the series ∑ k = 4 ∞ 1 (10 k - 9) ( ln (5 k + 8)) 3 can also be similar. In this rank, we can signify that the current viraz is also similar.
Proceed to prove ∑ k = 4 ∞ 1 (5 k + 8) (ln (5 k + 8)) 3 .
Scale function y = 1 5 x + 8 (ln (5 x + 8)) 3 greater than zero, do not break and change to [ 4 ; +∞). Vikoristovuemo sign, described in the front paragraph:
∫ 4 + ∞ d x (5 x + 8) (l n (5 x + 8)) 3 = lim A → + ∞ ∫ 4 A d x (5 x + 8) (ln (5 x + 8)) 3 = = 1 5 lim A → + ∞ ∫ 4 A d (ln (5 x + 8) (ln (5 x + 8)) 3 = - 1 10 lim A → + ∞ 1 (ln (5 x + 8)) 2 |4 A = = - 1 10 lim A → + ∞ 1 (ln (5 A + 8)) 2 - 1 (ln (5 4 + 8)) 2 = = - 1 10 1 + ∞ - 1 (ln 28) 2 = 1 10 ln 28 2
In the shortest series that converges, ∫ 4 + ∞ d x (5 x + 8) (ln (5 x + 8)) 3 , we can find that ∑ k = 4 ∞ 1 (5 k + 8) (ln (5 k + 8) )) 3 also converge.
Oznaka Raabe
It is possible that ∑ k = 1 ∞ a k is a positive sign number series.
Yakscho lim k → + ∞ k ak a k + 1< 1 , то ряд расходится, если lim k → + ∞ k · a k a k + 1 - 1 >1, then converge.
The Danish method of designation can be victorious in that case, as the described technique does not give visible results.
Doslіdzhennya on absolute zbіzhnіst
For the rest, we take ∑ k = 1 ∞ b k. Vikorist's positive ∑ k = 1 ∞ b k . We can vikoristovuvat be-yak z vіdpovіdnyh sign, yakі we described more. If the series ∑ k = 1 ∞ b k converge, then the original series is absolutely similar.
butt 18
Continue the series ∑ k = 1 ∞ (-1) k 3 k 3 + 2 k - 1 to the left ∑ k = 1 ∞ (- 1) k 3 k 3 + 2 k - 1 = ∑ k = 1 ∞ 1 3 k 3 + 2k-1.
Umov lim k → + ∞ 1 3 k 3 + 2 k - 1 = 1 + ∞ = 0 . Vikoristovo ∑ k = 1 ∞ 1 k 3 2 i speeds up with another sign: lim k → + ∞ 1 3 k 3 + 2 k - 1 1 k 3 2 = 1 3 .
The series ∑ k = 1 ∞ (-1) k 3 k 3 + 2 k - 1 converge. The outer row is also absolutely similar.
Razbіzhnіst znazmіnіh ryadі
Just as the series ∑ k = 1 ∞ b k is disparate, then the same familiar sign series ∑ k = 1 ∞ b k is either disparate or mentally similar.
Instead of d'Alembert's sign and the radical Cauchy's sign, it is possible to supplement the wisps about ∑ k = 1 ∞ b k for the expansion of modules ∑ k = 1 ∞ b k . The series ∑ k = 1 ∞ b k also diverge, so that the necessary mental feasibility does not win, so that lim k → ∞ + b k ≠ 0 .
butt 19
Reverse variability 1 7 , 2 7 2 , - 6 7 3 , 24 7 4 , 120 7 5 - 720 7 6 , . . . .
Module k-th member of representations ak b k = k! 7 k.
Continue the series ∑ k = 1 ∞ b k = ∑ k = 1 ∞ k! 7 k on the edge beyond the d'Alembert sign: lim k → + ∞ b k + 1 b k = lim k → + ∞ (k + 1) ! 7k + 1k! 7 k = 1 7 limk → + ∞ (k + 1) = + ∞.
∑ k = 1 ∞ b k = ∑ k = 1 ∞ k ! 7 k disperse like i, like i exit option.
butt 20
Chi є ∑ k = 1 ∞ (-1) k · k 2 + 1 ln (k + 1) similar.
Let's take a look at the necessary Umov's theory lim k → + ∞ b k = lim k → + ∞ k 2 + 1 ln (k + 1) = ∞ ∞ = lim k → + ∞ = k 2 + 1 "(ln (k + 1))" = = lim k → + ∞ 2 k 1 k + 1 = lim k → + ∞ 2 k (k + 1) = + ∞ . Umov is not a Vikonan, so ∑ k = 1 ∞ (- 1) k · k 2 + 1 ln (k + 1) is a series of expansions. The boundary of the bula was calculated according to the Lopital rule.
Signs of mental health
Sign of Leibnitz
Appointment 12As the size of the members of the series, which are drawn, change b 1 > b 2 > b 3 >. . . >. . . і inter module = 0 as k → + ∞ , then the series ∑ k = 1 ∞ b k runs.
butt 17
Take a look at ∑ k = 1 ∞ (-1) k 2 k + 1 5 k (k + 1) for the opportunity.
A series of representations yak ∑ k = 1 ∞ (-1) k 2 k + 1 5 k (k + 1) = ∑ k = 1 ∞ 2 k + 1 5 k (k + 1) . The need for umova lim k + ∞ = 2 k + 1 5 k (k + 1) = 0 . Let's look at ∑ k = 1 ∞ 1 k behind another equalization sign lim k → + ∞ 2 k + 1 5 k (k + 1) 1 k = lim k → + ∞ 2 k + 1 5 (k + 1) = 2 5
It is possible that ∑ k = 1 ∞ (-1) k 2 k + 1 5 k (k + 1) = ∑ k = 1 ∞ 2 k + 1 5 k (k + 1) diverge. Series ∑ k = 1 ∞ (- 1) k 2 k + 1 5 k (k + 1) converge after the Leibniz sign: sequence 2 1 + 1 5 1 1 1 + 1 = 3 10 , 2 2 + 1 5 2 (2 + 1) = 5 30 , 2 3 + 1 5 3 3 + 1, . . . changes i lim k → + ∞ = 2 k + 1 5 k (k + 1) = 0 .
A number of mentally converge.
Sign of Abel-Dirichlet
Appointment 13∑ k = 1 + ∞ u k · v k go off at that point, because ( u k ) is not growing, and the sequence ∑ k = 1 + ∞ v k is limited.
butt 17
Continue 1 - 3 2 + 2 3 + 1 4 - 3 5 + 1 3 + 1 7 - 3 8 + 2 9 +. . . for convenience.
visible
1 - 3 2 + 2 3 + 1 4 - 3 5 + 1 3 + 1 7 - 3 8 + 2 9 +. . . = 1 1 + 1 2 (- 3) + 1 3 2 + 1 4 1 + 1 5 (- 3) + 1 6 = ∑ k = 1 ∞ u k v k
de(u k) = 1, 1 2, 1 3,. . . - Unstable, and sequence (v k) = 1, - 3, 2, 1, - 3, 2,. . . fringed (S k ) = 1, - 2, 0, 1, - 2, 0,. . . . A number of converge.
How did you remember the pardon in the text, be kind, see it and press Ctrl + Enter
Harmony series- a sum, accumulated from an infinite number of members, wrapped in the last numbers of the natural series:
∑ k = 1 ∞ 1 k = 1 + 1 2 + 1 3 + 1 4 + ⋯ + 1 k + ⋯ )(k))=1+(\frac (1)(2))+(\frac (1) (3))+(\frac (1)(4))+cdots +(\frac (1) (k))+cdots ).Encyclopedic YouTube
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✪ Number rows. Basic understanding - bezbotvy
✪ Proof of the diversity of the harmonic series
✪ Number rows-9. Skhіdnіst i razbіzhnіst a number of Dirichle
✪ Consultation No. 1. Mat. analysis. Four's series according to the trigonometric system. The simplest power
✪ ROW. looking around
Subtitles
The sum of the first n terms in a row
Okremі members of the row to break zero, but yogo scrip to disperse. The n-th part sum of the s n harmonic series is called the n-th harmonic number:
s n = ∑ k = 1 n 1 k = 1 + 1 2 + 1 3 + 1 4 + ⋯ + 1 n (\displaystyle s_(n)=\sum _(k=1)^(n)(\frac (1 )(k))=1+(\frac (1)(2))+(\frac (1)(3))+(\frac (1)(4))+cdots +(\frac (1) ( n)))Actual values of private sums
| s 1 = 1 s 2 = 3 2 = 1 , 5 s 3 = 11 6 ≈ 1.833 s 4 = 25 12 ≈ 2.083 s 5 = 137 60 ≈ 2.283 (\displaystyle (\begin(matrix)s_(1)& \\ \\s_(2)&=&(\frac (3)(2))&=&1(,)5\\\\s_(3)&=&(\frac (11)(6))& \approx &1(,)833\\\\s_(4)&=&(\frac (25)(12))&\approx &2(,)083\\\s_(5)&=&(\frac (137) (60))&\approx &2(,)283\end(matrix))) | s 6 = 49 20 = 2.45 s 7 = 363 140 ≈ 2.593 s 8 = 761 280 ≈ 2.718 s 10 3 ≈ 7.484 s 10 6 ≈ 14.393 (\splaystyle(&be \frac (49)(20))&=&2 (,)45\\\\s_(7)&=&(\frac (363)(140))&\approx &2(,)593\\\s_ (8)&=&(\frac (761)( 280))&\approx &2(,)718\\\\s_(10^(3))&\approx &7(,)484\\\\s_( 10^(6))&\approx &14(,) 393\end(matrix))) |
Euler formula
At the value ε n → 0 (\displaystyle \varepsilon _(n)\rightarrow 0) father, for the great n (\displaystyle n):
s n ≈ ln (n) + γ (\displaystyle s_(n)\approx \ln(n)+\gamma )- Euler's formula for sumi first n (\displaystyle n) members of the harmonic series.| n (\displaystyle n) | s n = ∑ k = 1 n 1 k (\displaystyle s_(n)=\sum _(k=1)^(n)(\frac (1)(k))) | ln (n) + γ (\displaystyle \ln(n)+\gamma ) | ε n (\displaystyle \varepsilon _(n)), (%) |
| 10 | 2,93 | 2,88 | 1,7 |
| 25 | 3,82 | 3,80 | 0,5 |
The more exact asymptotic formula for the partial sum of the harmonic series is:
s n ≍ ln (n) + γ + 1 2 n − 1 12 n 2 + 1 120 n 4 − 1 252 n 6 ⋯ = ln (n) + γ + 1 2 n − ∑ k = 1 ∞ 2 k n 2 k (displaystyle s_(n)\asymp \ln(n)+\gamma +(\frac (1)(2n))-(\frac (1)(12n^(2)))+(\frac (1) (120n^(4)))-(\frac (1)(252n^(6)))\dots =\ln(n)+\gamma +(\frac (1)(2n))- \sum _( k=1)^(\infty )(\frac (B_(2k))(2k\,n^(2k)))), de B 2 k (\displaystyle B_(2k))- Numbers Bernoulli.Whose row to disperse, pro-pardon to calculate the new one in no way outweighs half of the first protruding member.
Number-theoretic power of private sums
∀ n > 1 s n ∉ N (\displaystyle \forall n>1\;\;\;\;s_(n)\notin \mathbb (N) )
Razbіzhnіst row
S n → ∞ (\displaystyle s_(n)\rightarrow \infty ) at n → ∞ (\displaystyle n\rightarrow \infty )
Harmonious series to diverge even better (in order for the private sum to exceed 100, it is necessary to have close to 1043 elements in a row).
The diversity of the harmonic row can be demonstrated by aligning it with the telescopic row:
v n = ln (n + 1) − ln n = ln (1 + 1 n) ~ + ∞ 1 n (\displaystyle v_(n)=\ln(n+1)-\ln n=\ln \ left(1+(\frac (1)(n))\right)(\underset (+\infty )(\sim ))(\frac (1)(n))),chastkov's sum of which, obviously, is more expensive:
∑ i = 1 n − 1 v i = ln n ~ s n (\displaystyle \sum _(i=1)^(n-1)v_(i)=\ln n\sim s_(n)).Orem's proof
Proof of diversity can be encouraged, grouping dodanki in this order:
∑ k = 1 ∞ 1 k = 1 + [ 1 2 ] + [ 1 3 + 1 4 ] + [ 1 5 + 1 6 + 1 7 + 1 8 ] + [ 1 9 + ⋯ ] + ⋯ > 1 + [ 1 2 ] + [ 1 4 + 1 4 ] + [ 1 8 + 1 8 + 1 8 + 1 8 ] + [ 1 16 + ⋯ ] + ⋯ = 1 + 1 2 + 1 2 + 1 2 + 1 2 + ⋯ . (\displaystyle (\begin(aligned)\sum _(k=1)^(\infty )(\frac (1)(k))&()=1+\left[(\frac (1)(2) )\right]+\left[(\frac (1)(3))+(\frac (1)(4))\right]+\left[(\frac (1)(5))+(\frac (1)(6))+(\frac (1)(7))+(\frac (1)(8))\right]+\left[(\frac (1)(9))+\cdots \ right]+\cdots \\&()>1+\left[(\frac (1)(2))\right]+\left[(\frac (1)(4))+(\frac (1) (4))\right]+\left[(\frac (1)(8))+(\frac (1)(8))+(\frac (1)(8))+(\frac (1) (8))\right]+\left[(\frac (1)(16))+cdots \right]+cdots \&()=1+\ (\frac (1)(2))\ \ \ + \quad (\frac (1) (2)) \ \quad + \qquad \quad (\frac (1) (2)) \quad \ \quad \ + \quad \ (\frac (1)(2)) \quad +\cdots .end(aligned)))The rest of the row, obviously, to disperse. This proof belongs to the middle-class clergyman Mykola Orem (bl. 1350).
Alternative proof of dissimilarity
proponuєmo chitachevi perekonatisya at the pardon of that proof
Retail n (\displaystyle n)-th harmonic number and natural logarithm n (\displaystyle n) converge to the post-Euler-Mascheroni.
The difference between different harmonic numbers is in no way equal to the whole number and the same harmonic number, crim H 1 = 1 (\displaystyle H_(1)=1) not є qіlim.
Tied rows
Row Dirichle
Let us name the row
∑ k = 1 ∞ 1 k α = 1 + 1 2 α + 1 3 α + 1 4 α + ⋯ + 1 k α + ⋯ (\displaystyle \sum _(k=1)^(\infty )(\frac ( 1)(k^(\alpha )))=1+(\frac (1)(2^(\alpha )))+(\frac (1)(3^(\alpha )))+(\frac ( 1)(4^(\alpha )))+\cdots +(\frac (1)(k^(\alpha )))+\cdots ).The harmonization series diverge when α ⩽ 1 (\displaystyle \alpha \leqslant 1) i converge for α > 1 (\displaystyle \alpha >1) .
The sum of the knotted harmonic row in order α (\displaystyle \alpha) more important value of the Riemann zeta function:
∑ k = 1? ))For guys, the value is clearly expressed through the number pi, for example, ζ (2) = π 2 6 (\displaystyle \zeta (2)=(\frac (\pi ^(2))(6))), And for α=3 its value is analytically unknown.
Another illustration of the diversity of the harmonic series can be ζ (1 + 1 n) ~ n (\displaystyle \zeta (1+(\frac (1)(n)))\sim n) . To this it seems that such a series can be imovirnist 1, and in the sum of the series it is a vipadkovy value with powerful powers. For example, the function, power, and mobility, calculated in points +2 or −2 may be the value:
0,124 999 999 999 999 999 999 999 999 999 999 999 999 999 7 642 …,ventilated in ⅛ less lower 10 −42 .
"Vitonic" harmony series
Kempner series (English)If you look at the harmony row, in which only dodanki are missing, the banners of which do not avenge the number 9, then it will appear, what is the sum, what is lost, converge to the number<80 . Более того, доказано, что если оставить слагаемые, не содержащие любой заранее выбранной последовательности цифр, то полученный ряд будет сходиться. Однако из этого будет ошибочно заключать о сходимости исходного гармонического ряда, так как с ростом разрядов в числе n (\displaystyle n), less and less dodankiv is taken for the sumi of the “wound” series. So that in the final rahunka it is seen that the greater number of members is more important, so that the sum of the harmonic series is established, so as not to overturn the geometric progression that surrounds the beast.
Pereviriti zbіzhnist a number of ways can kіlkom. First, you can just know sum in a row. As a result, we take the final number, then such series converge. For example, shards
then the series will converge. Even though we were not able to know the sum of the series, we should try to win other methods for rechecking the profitability of the series.
One of these methods is d'Alembert sign
here i must match the n-th and (n+1)-th members of the series, and the value depends on the values of D: Like D< 1 - ряд сходится, если D >
Like a butt, doslіdzhuєmo zbіzhnіst is low with additional signs of d'Alembert. We will write down the verse for i. Now we know vіdpovіdny boundary :
Shards, visible to the signs of d'Alembert, the series converge.
Another method that allows you to reverse the number of radical sign Kosha, which is recorded by the coming rank:
here is the n-th member of the series, and zbіzhnist, as in the case of d'Alembert's signs, is assigned to the values of D: Like D< 1 - ряд сходится, если D >1 - disperse. When D = 1 - this sign does not give any additional evidence and it is necessary to carry out additional follow-up.
As an example, doslіdzhuєmo zbіzhnіst a number of additional radical signs of Koshі. Let's write down a viraz for . Now we know the difference between:
Oskіlki vydpovіdno to radical signs of Koshі, a number of disperse.
Varto designate, scho the order with re-arrangement, to use other signs of the zbіzhnostі rows, such as the integral sign of Kosh, the sign of Raabe and іn.
Our online calculator, Inducements on the basis of the Wolfram Alpha system allows you to protest in a row. When tsiomu, like a calculator for a series, it looks like a specific number, the series converges. Otherwise, it is necessary to pay attention to the paragraph "Test of the performance in a row." If there is a phrase “series converges”, then the series converges. If there is a phrase “series diverges”, then the series will diverge.
Below is a translation of the possible values for the item “Row performance test”:
| Text on English language | Russian text |
|---|---|
| By harmonic series test, the series divergences. | When por_vnyannі doslіdzhuvannogo row with harmonіymi row vihіdny row disperse. |
| ratio test is inconclusive. | The sign of d'Alembert is impossible to date in a row. |
| The root test is inconclusive. | The radical sign of Kosh can not give any indications about the success of the series. |
| By comparison test, the series converges. | Behind the sign of alignment, the series converge |
| By ratio test, the series converges. | Behind the d'Alembert sign, the series converge |
| By the limit test, the series divergences. | Based on the fact that title="(!LANG:(!LANG:Between the nth member of the series when n->oo is not equal to zero or not"> , или указанный предел не существует, сделан вывод о том, что ряд расходится. !}!} |
Vidpovid: row to diverge.
Butt #3
Find the sum of the series $\sum\limits_(n=1)^(\infty)\frac(2)((2n+1)(2n+3))$.
The oskіlki lower intera pіdsumovuvannya dorіvnyuє 1, then the main member of the series of entries under the sumi sign: $u_n=\frac(2)((2n+1)(2n+3))$. Storing the nth private sum is low, tobto. supposedly the first $n$ members of the given numerical series:
$$ S_n=u_1+u_2+u_3+u_4+\ldots+u_n=\frac(2)(3\cdot 5)+\frac(2)(5\cdot 7)+\frac(2)(7\cdot 9 )+\frac(2)(9cdot 11)+ldots+frac(2)((2n+1)(2n+3)). $$
Why I write $\frac(2)(3\cdot 5)$ myself, and not $\frac(2)(15)$, will be clear from afar. However, the record of the private sum did not bring us one iota closer. Even if we need to know $\lim_(n\to\infty)S_n$, otherwise we can simply write:
$$ \lim_(n\to\infty)S_n=\lim_(n\to\infty)\left(\frac(2)(3\cdot 5)+\frac(2)(5\cdot 7)+\ frac(2)(7\cdot 9)+\frac(2)(9\cdot 11)+\ldots+\frac(2)((2n+1)(2n+3))\right), $$
then this record, absolutely true to form, will not give us anything in essence. Schob to know the boundary, viraz the private sumi, it is necessary to ask ahead.
For which is the standard transformation, which is used in the fraction $\frac(2)((2n+1)(2n+3))$, as it represents the main member of the series, elementary fractions. The food distribution of rational fractions on the elementary is dedicated to the topic (div., for example, butt No. 3 on the other side). Expanding $\frac(2)((2n+1)(2n+3))$ into elementary fractions, math:
$$ \frac(2)((2n+1)(2n+3))=\frac(A)(2n+1)+\frac(B)(2n+3)=\frac(A\cdot(2n ) +3)+B\cdot(2n+1))((2n+1)(2n+3)). $$
Let's compare the numbers of fractions in the left and right parts of the taken equality:
$$ 2=Acdot(2n+3)+Bcdot(2n+1). $$
To know the value of $A$ and $B$, є two ways. You can open the arms and regroup the dodanki, or you can just put in the replacement of $n$ acts valid values. For example, for diversity in the first butt, it is given the first way, and for the next one - it is given the private value of $n$. Opening the arches and regrouping the dodanki, it’s necessary:
$$ 2=2An+3A+2Bn+B; 2=(2A+2B)n+3A+B. $$
The left part of equality has zero before $n$. As always, the last part of equanimity for accuracy is possible as $0\cdot n+ 2$. Since the left part of equality has zero in front of $n$, and the right part of equality has $2A+2B$ in front of $n$, then perhaps the first equal: $2A+2B=0$. Once again, we divide the offending part of this equal by 2, subtracting the last $A+B=0$.
The pieces of the left part of the equality of the equal term are equal to 2, and the right part of the equality of the equal member of the equal length is $3A+B$, then $3A+B=2$. Otzhe, maєmo system:
$$ \left\(\begin(aligned) & A+B=0;\\ & 3A+B=2. \end(aligned)\right. $$
The proof is carried out by the method of mathematical induction. On the first step, it is necessary to reconsider, what is meant by equality, that $S_n=\frac(1)(3)-\frac(1)(2n+3)$ for $n=1$. We know that $S_1=u_1=\frac(2)(15)$, but we would like to give $\frac(1)(3)-\frac(1)(2n+3)$ the value of $\frac(2 ) (15) $, how to put in new $ n = 1 $? Revisited:
$$ \frac(1)(3)-\frac(1)(2n+3)=\frac(1)(3)-\frac(1)(2\cdot 1+3)=\frac(1) (3)-frac(1)(5)=frac(5-3)(15)=frac(2)(15). $$
Also, for $n=1$, $S_n=\frac(1)(3)-\frac(1)(2n+3)$ is equal. For whom the first step to the method of mathematical induction is completed.
It is acceptable that $n=k$ equality is vikonano, that is. $S_k=\frac(1)(3)-\frac(1)(2k+3)$. Let's say that this equanimity will be won for $n=k+1$. For which $S_(k+1)$ can be considered:
$$ S_(k+1)=S_k+u_(k+1). $$
$u_n=\frac(1)(2n+1)-\frac(1)(2n+3)$, then $u_(k+1)=\frac(1)(2(k+1)+ 1 )-frac(1)(2(k+1)+3)=frac(1)(2k+3)-frac(1)(2(k+1)+3)$. It is evident that $S_k=\frac(1)(3)-\frac(1)(2k+3)$ has been stretched to the point of crushing, so the formula $S_(k+1)=S_k+u_(k+1)$ will be seen :
$$ S_(k+1)=S_k+u_(k+1)=\frac(1)(3)-\frac(1)(2k+3)+\frac(1)(2k+3)-\ frac(1)(2(k+1)+3)=frac(1)(3)-frac(1)(2(k+1)+3). $$
Visnovok: the formula $S_n=\frac(1)(3)-\frac(1)(2n+3)$ is valid for $n=k+1$. Also, using the method of mathematical induction, the formula $S_n=\frac(1)(3)-\frac(1)(2n+3)$ is true for any $n\in N$. Equity has been brought.
At the standard course advanced mathematics sound satisfied with the "rejoicing" of the dodankiv, which are soon, not depending on the daily proofs. Later, we took away the viraz for n-ї private sumi: $ S_n = frac (1) (3) - frac (1) (2n + 3) $. We know the value of $\lim_(n\to\infty)S_n$:
Visnovok: the number of tasks converges i th sum $S=\frac(1)(3)$.
Another way to simplify the formula for a private sum.
To be honest, apparently, I myself see the difference in the same way :) Let's write down the private sum in a quick version:
$$ S_n=\sum\limits_(k=1)^(n)u_k=\sum\limits_(k=1)^(n)\frac(2)((2k+1)(2k+3)). $$
We took earlier that $u_k=\frac(1)(2k+1)-\frac(1)(2k+3)$ to:
$$ S_n=\sum\limits_(k=1)^(n)\frac(2)((2k+1)(2k+3))=\sum\limits_(k=1)^(n)\left (\frac(1)(2k+1)-\frac(1)(2k+3)\right). $$
Sum of $S_n$ to avenge the kіlkіst kіlkіst of dodankіv, so we can rearrange them in such a way as we are tempted. I want to fold all the warehouses in the form $\frac(1)(2k+1)$, and then move on to the bottom in the form $\frac(1)(2k+3)$. Tse means that we give a private sum to such a person:
$$ S_n =\frac(1)(3)-\frac(1)(5)+\frac(1)(5)-\frac(1)(7)+\frac(1)(7)-\ frac(1)(9)+frac(1)(9)-frac(1)(11)+ldots+frac(1)(2n+1)-frac(1)(2n+3)= \\ =\ frac(1)(3)+\frac(1)(5)+\frac(1)(7)+\frac(1)(9)+\ldots+\frac(1)(2n+1 )-\left (\frac(1)(5)+\frac(1)(7)+\frac(1)(9)+\ldots+\frac(1)(2n+3)\right). $$
Obviously, the open record is not handy, so the presented equality can be made more compact:
$$ S_n=\sum\limits_(k=1)^(n)\left(\frac(1)(2k+1)-\frac(1)(2k+3)\right)=\sum\limits_( k=1)^(n)\frac(1)(2k+1)-\sum\limits_(k=1)^(n)\frac(1)(2k+3). $$
Now we can convert $\frac(1)(2k+1)$ and $\frac(1)(2k+3)$ to one form. I vvazhim zruchny bring to the sight of a larger fraction (if you can and to a lesser, tse relish on the right). So, like $\frac(1)(2k+1)>\frac(1)(2k+3)$ (the larger the banner, the smaller the drіb), then we induce drіb $\frac(1)(2k+3) $ looks like $\frac(1)(2k+1)$.
Viraz at the banner of the fraction $\frac(1)(2k+3)$ I will present it in this way:
$$ \frac(1)(2k+3)=frac(1)(2k+2+1)=frac(1)(2(k+1)+1). $$
The sum $\sum\limits_(k=1)^(n)\frac(1)(2k+3)$ can now be written like this:
$$ \sum\limits_(k=1)^(n)\frac(1)(2k+3)=\sum\limits_(k=1)^(n)\frac(1)(2(k+1 ) )+1)=\sum\limits_(k=2)^(n+1)\frac(1)(2k+1). $$
How equal $\sum\limits_(k=1)^(n)\frac(1)(2k+3)=\sum\limits_(k=2)^(n+1)\frac(1)(2k+ 1 ) $ do not call the food, then send a demo. If there is food, then I ask you to spread the note.
How did we take away the converted bag? show/hide
We have a series $\sum\limits_(k=1)^(n)\frac(1)(2k+3)=\sum\limits_(k=1)^(n)\frac(1)(2( k+1)+1)$. Let's change $k+1$ and introduce a new change, for example $t$. Also, $ t = k + 1 $.
How did the old $k$ change change? And it changed from 1 to $ n $. Let's find out how the new $t$ will be changed. If $k=1$, then $t=1+1=2$. If $k=n$, then $t=n+1$. Later, viraz $\sum\limits_(k=1)^(n)\frac(1)(2(k+1)+1)$ now becoming $\sum\limits_(t=2)^(n +1)\frac(1)(2t+1)$.
$$ \sum\limits_(k=1)^(n)\frac(1)(2(k+1)+1)=\sum\limits_(t=2)^(n+1)\frac(1 )(2t+1). $$
We have є sum $\sum\limits_(t=2)^(n+1)\frac(1)(2t+1)$. Nutrition: but what is not the same, how can I beat the letter in my sum? :) Tritely writing down the letter $k$ instead of $t$, take the step forward:
$$ \sum\limits_(t=2)^(n+1)\frac(1)(2t+1)=\sum\limits_(k=2)^(n+1)\frac(1)(2k +1). $$
Return $\sum\limits_(k=1)^(n)\frac(1)(2(k+1)+1)=\sum\limits_(k=2)^(n+1) \ frac(1)(2k+1)$.
In this rank, a private sum can be paid from such a look:
$$ S_n=\sum\limits_(k=1)^(n)\frac(1)(2k+1)-\sum\limits_(k=1)^(n)\frac(1)(2k+3 )=\sum\limits_(k=1)^(n)\frac(1)(2k+1)-\sum\limits_(k=2)^(n+1)\frac(1)(2k+1 ). $$
Respect that sumi $\sum\limits_(k=1)^(n)\frac(1)(2k+1)$ i $\sum\limits_(k=2)^(n+1)\frac(1 )(2k+1)$ Zrobimo qi between the same. "Taking" the first element from sum $\sum\limits_(k=1)^(n)\frac(1)(2k+1)$ will be:
$$ \sum\limits_(k=1)^(n)\frac(1)(2k+1)=\frac(1)(2\cdot 1+1)+\sum\limits_(k=2)^ (n)\frac(1)(2k+1)=\frac(1)(3)+\sum\limits_(k=2)^(n)\frac(1)(2k+1). $$
"Taking" the remaining element from the sum $\sum\limits_(k=2)^(n+1)\frac(1)(2k+1)$, we can take:
$$\sum\limits_(k=2)^(n+1)\frac(1)(2k+1)=\sum\limits_(k=2)^(n)\frac(1)(2k+1 )+\frac(1)(2(n+1)+1)=\sum\limits_(k=2)^(n)\frac(1)(2k+1)+\frac(1)(2n+ 3 ).$$
Todi viraz for the private sumi in the future I look:
$$ S_n=\sum\limits_(k=1)^(n)\frac(1)(2k+1)-\sum\limits_(k=2)^(n+1)\frac(1)(2k +1)=\frac(1)(3)+\sum\limits_(k=2)^(n)\frac(1)(2k+1)-\left(\sum\limits_(k=2)^ (n)\frac(1)(2k+1)+\frac(1)(2n+3)\right)=\\ =\frac(1)(3)+\sum\limits_(k=2)^ (n)\frac(1)(2k+1)-\sum\limits_(k=2)^(n)\frac(1)(2k+1)-\frac(1)(2n+3)=\ frac(1)(3)-\frac(1)(2n+3). $$
If you skip all the explanations, then the process of calculating the short formula for the n-ї private sum should look like this:
$$ S_n=\sum\limits_(k=1)^(n)u_k =\sum\limits_(k=1)^(n)\frac(2)((2k+1)(2k+3)) = \sum\limits_(k=1)^(n)\left(\frac(1)(2k+1)-\frac(1)(2k+3)\right)=\\ =\sum\limits_(k =1)^(n)\frac(1)(2k+1)-\sum\limits_(k=1)^(n)\frac(1)(2k+3) =\frac(1)(3) +\sum\limits_(k=2)^(n)\frac(1)(2k+1)-\left(\sum\limits_(k=2)^(n)\frac(1)(2k+1 ) )+frac(1)(2n+3)right)=frac(1)(3)-frac(1)(2n+3). $$
Guess what we did to make $frac(1)(2k+3)$ look like $frac(1)(2k+1)$. Zrozumilo, you can and navpaki, tobto. reveal drіb $\frac(1)(2k+1)$ like $\frac(1)(2k+3)$. Kіntsevy viraz for private sumi does not change. The process of znakhodzhennya chastkovoї sumi in tsomu vipadku I prihovayu pіd primіtku.
How to know $S_n$, how to reduce a fraction to look different? show/hide
$$ S_n =\sum\limits_(k=1)^(n)\frac(1)(2k+1)-\sum\limits_(k=1)^(n)\frac(1)(2k+3 ) =\sum\limits_(k=0)^(n-1)\frac(1)(2k+3)-\sum\limits_(k=1)^(n)\frac(1)(2k+3 )=\\ =\frac(1)(3)+\sum\limits_(k=1)^(n-1)\frac(1)(2k+3)-\left(\sum\limits_(k= 1)^(n-1)\frac(1)(2k+3)+\frac(1)(2n+3)\right) =\frac(1)(3)-\frac(1)(2n+ 3 ). $$
Also, $S_n=\frac(1)(3)-\frac(1)(2n+3)$. We know between $\lim_(n\to\infty)S_n$:
$$ \lim_(n\to\infty)S_n=\lim_(n\to\infty)\left(\frac(1)(3)-\frac(1)(2n+3)\right)=\frac (1)(3)-0=\frac(1)(3). $$
The number of tasks converges i th sum $S=\frac(1)(3)$.
Vidpovid: $S=\frac(1)(3)$.
Prodovzhennya those znakhodzhennya sumi row will be looked at in other and third parts.
Enthusiasm...