The problem states:

Wherever it converges, find the exact sum of


Using ratio test I know the series converges iff $|x|>1$, but have not idea how to calculate the sum.

Any help is highly appreciated. Thanks and regards.

  • 2
    $\begingroup$ Does it look more familiar if you substitute $x = 1/y$ ? $\endgroup$ – Martin R Mar 1 '15 at 21:15
  • $\begingroup$ @MartinR Definitely. I'm feeling a little embarrased... $\endgroup$ – Ab urbe condita Mar 1 '15 at 21:29

Follow Martin R's comment. Keep in mind that if the function is uniformly continuous, you can interchange infinite sum and differentiation and also $ky^k = y \frac{dy^k}{dy}$.


Let $S = \sum_{n = 1}^\infty na^n$, where $a = 1/x$. Then

$$aS = \sum_{n = 1}^\infty na^{n+1} = \sum_{n = 1}^\infty (n + 1)a^{n+1} - \sum_{n = 1}^\infty a^{n+1} = S - a - \frac{a^2}{1 - a} = S - \frac{a}{1 - a}.$$

Thus $(1 - a)S = a/(1 - a)$, yielding

$$S = \frac{a}{(1 - a)^2} = \frac{\frac{1}{x}}{\left(1 - \frac{1}{x}\right)^2} = \frac{x}{(x - 1)^2}.$$


Here is a useful finite evaluation: $$ 1+r+r^2+...+r^n=\frac{1-r^{n+1}}{1-r}, \quad |r|<1. \tag1 $$ Then by differentiating $(1)$ you get $$ 1+2r+3r^2+...+nr^{n-1}=\frac{1-r^{n+1}}{(1-r)^2}+\frac{-(n+1)r^{n}}{1-r}, \quad |r|<1, \tag2 $$ and by making $n \to +\infty$ in $(2)$, using $|r|<1$, gives

$$ 1+2r+3r^2+...+nr^{n-1}+...=\frac{1}{(1-r)^2} \tag3 $$

If you multiply $(3)$ by $r$ and set $r=\dfrac1x$, you obtain an answer to your question.


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