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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.

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2 Answers 2

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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.

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    $\begingroup$ Need to know the series converges before you are allowed to do that. $\endgroup$
    – fleablood
    Commented Dec 13, 2017 at 6:51
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$f(x) = \sum_{k=1}^{n} x^k.$

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

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

Does it converge, what is the limit?

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