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$$\sum_{k=0}^\infty \frac{k+1}{3^k}$$

Steps:

$$\sum_{k=0}^\infty \frac{k}{3^k} + \sum_{k=1}^\infty \frac{1}{3^k}$$

$$Sn = \frac {a1(1-r^n)}{1-r}$$

$$r = \frac{second term}{first term}$$ $$a1 = \frac {1}{3^0}$$ $$a2 = \frac {1}{3} + \frac {1}{3^1}$$

$a1 = 0+1 = 1$
a2 = $\frac{2}{3}$

r = 2/3

$$S∞ = \frac {a1(1-r^∞)}{1-r}$$ $$S∞ = \frac {1(1- (2/3)^∞)}{1-(2/3)}$$ $$S∞ = \frac {1(1- 0)}{1-(2/3)}$$ $$S∞ = \frac {1}{1/3}$$ $$S∞ = 3$$

Here I get the sum being equal to three when in fact it is apparently 9/4. I am finding very limited information on how to solve infinite sum of series.

jimjim
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sawreals2
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3 Answers3

2

hint:

$$\sum_{k=0}^n(k+1)x^k=\biggl(\sum_{k=0}^nx^{k+1}\biggr)'=\biggl(\sum_{k=1}^{n+1}x^{k}\biggr)'.$$

Some details:

As a result,$$\sum_{k=0}^\infty(k+1)x^k=\biggl(\sum_{k=1}^{\infty}x^{k}\biggr)'.$$ Now it is well-known that $$\sum_{k=1}^{\infty}x^{k}=\frac x{1-x}=-1+\frac1{1-x}, \enspace\text{so }\sum_{k=0}^\infty(k+1)x^k=\\frac1{(1-x)^2}.$$

Bernard
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  • what is x here? – sawreals2 Feb 13 '18 at 00:11
  • Any number. For instance $\frac13$… ;o) – Bernard Feb 13 '18 at 00:15
  • Can you really use any number? If I wanted to use $$\frac {45}{6}$$ would I still get the right answer? – sawreals2 Feb 13 '18 at 00:16
  • As far as it is a finite sum, yes. If you have to go to infinity, it requires $|x|<1$. B.t.w., I've modified the hint, since you've changed the initial value of $k$. – Bernard Feb 13 '18 at 00:21
  • So if it is an infinite sum, x can be any number greater than 0 and less than 1. Does your steps to solving an infinite sum of series hold true for solving any infinite sum of series? I can use those steps for finding any sum of finite or infinite series? Also, why does k = 0 increase to k=1 in the last part of that equation. – sawreals2 Feb 13 '18 at 00:24
  • The property I used is that the sum of a (convergent) power series can be differentiated term by term (it also can be integrated), with the same radius of convergence – Bernard Feb 13 '18 at 00:27
  • Using your answer I am now getting 1 for the sum of series. $$\biggl(\sum_{k=1}^{n+1}x^{k}\biggr)'.$$ taking the derivative of $x^k$ is = $k(x)^0$ this is equal to 1 when k=1. – sawreals2 Feb 13 '18 at 00:31
  • You first must use the formula for the sum of the geometric series. – Bernard Feb 13 '18 at 00:34
  • Ok my original question shows that is equal to 3. – sawreals2 Feb 13 '18 at 00:36
  • I have now tried $S = \frac{a1}{1-r}$ where r = the common ratio between a2/a1, and again I am getting the sum = 3. – sawreals2 Feb 13 '18 at 00:50
  • With the way I point to, you should find $9/4$. – Bernard Feb 13 '18 at 00:54
  • Using any portion of that equation will yield the same result, as all parts are equal to eachother. I will use $$\sum_{k=0}^n(k+1)x^k$$ , this results in $$\sum_{k=0}^n(0+1)(1/3)^0$$ = 1(1) = 1. Additionally I am unsure of how to take a derivative of a sum of series. – sawreals2 Feb 13 '18 at 01:00
  • Which exponent is respective of the entire term? Is it 1 or k? k is the exponent of the term x. It would make sense that 1 is the term of everything in brackets then. However, if that is the case then 1-1 = 0, making the exponent of the sum of series 0 and thus = 1. This would mean that the derivative of the sum of series if the third part of the equation is also 1. – sawreals2 Feb 13 '18 at 01:10
  • No in you last but one comment, you did not calculate the sum of the series: $k$ has to be increased by $1$ at each step. Also, to differentatiate the sum of the series, you simply have to differentiate its closed form. – Bernard Feb 13 '18 at 01:15
  • And how do you find the closed form of an infinite sum of series? – sawreals2 Feb 13 '18 at 01:18
  • It's the geometric series! And it's not an infinite sum of series, it's the series which is an infinite sum (of powers of $x$ in the present case). – Bernard Feb 13 '18 at 01:25
  • Yes, that is what I meant. Geometric series with infinite sum. I have found the first four terms (I am unsure of how many terms I need to evaluate). The first four terms are: 1, 2/3, 1/3, and 4/27. I did this because you said in your comment you need to increase k by 1 at each step to calculate the sum of series. However I think it would be impossible to do this infinitely many times. This is where I am unsure of how to proceed. – sawreals2 Feb 13 '18 at 01:31
  • The common ratio here is not the same between each set of successive terms. For example between the first two terms it is 2/3, and between the next set term a3 and a2 it is 2. – sawreals2 Feb 13 '18 at 01:49
  • Actually the ratio between term three and term 2 is 1/2, but that does not detract from my point. In order for this to be a geometric series the common ratio MUST be the same for every successive term. Here this is not the case. – sawreals2 Feb 13 '18 at 01:57
  • It is the derivative of the geometric series, it's not the geometric series itself. – Bernard Feb 13 '18 at 10:01
  • Yes I have now solved this problem for myself. What I needed to do to visualize this problem was separate it into power series and geometric series: sum of k/3^k + sum of 1/3^k. – sawreals2 Feb 13 '18 at 10:15
  • That's a more complicated way, in my opinion. – Bernard Feb 13 '18 at 10:18
  • I've added details for the general formula. – Bernard Feb 13 '18 at 10:25
  • It appears quite similar to what I did, the geometric series sum is quite simple. You differentiated, after that substitute in $$\frac {1}{3}$$ for x. You say it is simpler to do it this way, why? I also see why you cannot take k=0 of derivative of the sum of series, and instead k=1. – sawreals2 Feb 13 '18 at 10:37
  • The's no $k=0$ in the r.h.s. because i won't appear anyway! For you your way, how do you manage $\sum\frac k{3^k}$? – Bernard Feb 13 '18 at 10:41
1

$$\begin{align} \sum_{k=0}^\infty \frac{k+1}{3^k} &=\frac{3}{2}\bigg(1-\frac{1}{3}\bigg)\sum_{k=0}^\infty \frac{k+1}{3^k}\\ &=\frac{3}{2}\sum_{k=0}^\infty \bigg(\frac{k+1}{3^k}-\frac{k+1}{3^{k+1}}\bigg)\\ &=\frac{3}{2}\bigg(\frac{1}{3^0}-\frac{1}{3^1}+\frac{2}{3^1}-\frac{2}{3^{2}}+\frac{3}{3^2}-\frac{3}{3^3}+\frac{4}{3^3}-\frac{4}{3^4}+...\bigg)\\ &=\frac{3}{2}\bigg(\frac{1}{3^0}+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...\bigg)\\ &=\frac{3}{2}\bigg(\frac{1}{1-\frac{1}{3}}\bigg)\\ &=\frac{3}{2}\cdot \frac{3}{2} \\ &=\frac{9}{4} \\ \end{align}$$

Franklin Pezzuti Dyer
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HINT

In general, $$ S(x) = \sum_{k=0}^\infty x^k = \frac{1}{1-x}, \quad \forall |x| < 1. $$ Therefore, $$ S'(x) = \frac{d}{dx}\sum_{k=0}^\infty x^k = \sum_{k=0}^\infty \frac{d}{dx}x^k = \sum_{k=1}^\infty k x^{k-1} = \frac{1}{x} \sum_{k=1}^\infty k x^k $$ but transforming the other side, $$ S'(x) = \frac{d}{dx} \frac{1}{1-x} = \frac{1}{(1-x)^2}. $$ Therefore, $$ \frac{x}{(1-x)^2} = xS'(x) = \sum_{k=1}^\infty k x^k, \quad \forall |x| <1. $$

gt6989b
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