How to find the infinite sum

Question

I need to find the infinite sum of the following series expansion

$$1/3 + 2/3^2 + 3/3^3 + 4/3^4 + \dots + k/3^k + \dots$$

I know that

$$x/(1 - x) = x + x^2 + x^3 + \dots + x^k + \dots$$

We need to find the $x$ value in order to find the infinite sum. What could the $x$ value be? I am not sure.

See https://math.stackexchange.com/questions/647587/sum-of-a-power-series-n-xn — lab bhattacharjee, Dec 13 '17 at 06:14

score 3 · Answer 1 · answered Dec 13 '17 at 06:19

3

Note that we have, $$S = \frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3}+ \ldots\tag{1}$$ $$\frac{1}{3} S = \frac{1}{3^2}+\frac{2}{3^3}+\ldots\tag{2}$$

Subtract the two equations and use the formula you have mentioned.

answered Dec 13 '17 at 06:19

1

Need to know the series converges before you are allowed to do that. – fleablood Dec 13 '17 at 06:51

score 0 · Answer 2 · answered Dec 13 '17 at 07:19

0

$f(x) = \sum_{k=1}^{n} x^k.$

Geometric series converges for $|x|\lt 1.$

$xf'(x)= \sum_{k=1}^{n}kx^k.$

Does it converge, what is the limit?

answered Dec 13 '17 at 07:19

Peter Szilas

2 Answers2