0
$\begingroup$

The problem states:

Wherever it converges, find the exact sum of

$$\displaystyle\sum_{n=1}^{\infty}\frac{n}{x^n}$$

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.

$\endgroup$
  • 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
3
$\begingroup$

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

$\endgroup$
4
$\begingroup$

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

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.