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I am interested in

$$\sum_{k=1}^{\infty} ke^{-k}$$

And I can get the closed form on Wolfram Alpha, $\frac{e}{(e-1)^2}$, but I am curious how to derive it. It doesn't appear to be a typical geometric series.

user525966
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    Solve $\sum e^{-k}$ and then take derivatives. – user295959 Feb 25 '18 at 19:42
  • Here is one method you can try. First compute $f(x) = \sum_{k=0}^\infty x^k$ (geometrical series) and then show that $x f'(x) = \sum_{k=0}^\infty k x^k$ and finally you just plug in $x= e^{-1}$. – Winther Feb 25 '18 at 19:44

4 Answers4

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In fact it is the derivative of a geometric series because $$ \left(e^{-kx}\right)'=-ke^{-kx} $$ You can for example calculate for $x \in \mathbb{N}^{*}$ $$ \sum_{k=0}^{+\infty}e^{-kx}=\frac{1}{1-e^{-x}} $$ Then differenciating $$ \sum_{k=0}^{+\infty}ke^{-kx}=-\left(\frac{1}{1-e^{-x}}\right)'=\frac{e^x}{\left(e^x-1\right)^2} $$ Taking $x=1$ gives you the expected result.

Atmos
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There are many ways to evaluate $S = \sum_{k=1}^{\infty} kx^k $.

This is possibly the simplest.

$xS = x\sum_{k=1}^{\infty} kx^k = \sum_{k=1}^{\infty} kx^{k+1} = \sum_{k=2}^{\infty} (k-1)x^{k} $ so

$\begin{array}\\ S-xS &= \sum_{k=1}^{\infty} kx^k-\sum_{k=2}^{\infty} (k-1)x^{k}\\ &= x+\sum_{k=2}^{\infty} kx^k-\sum_{k=2}^{\infty} (k-1)x^{k}\\ &= x+\sum_{k=2}^{\infty} (kx^k-(k-1)x^{k})\\ &= x+\sum_{k=2}^{\infty} x^k\\ &= x+\dfrac{x^2}{1-x}\\ &= \dfrac{x-x^2+x^2}{1-x}\\ &= \dfrac{x}{1-x}\\ \text{so}\\ S &= \dfrac{x}{(1-x)^2}\\ \end{array} $

Then put $x = e^{-1}$.

marty cohen
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For each real $x $, $e^{-x}<1$ and

$$e^{-x}+e^{-2x}+....=\frac {1}{1-e^{-x}} $$

differentiation gives

$$-(e^{-x}+2e^{-2x}+....)=\frac {-e^{-x}}{(1-e^{-x})^2} $$ $$=\frac {-e^x}{(e^x-1)^2} $$

For $x=1$, you get the result.

hamam_Abdallah
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By elementary means:

$$S:=a+a^2+a^2+a^3+a^3+a^3+a^4+a^4+a^4+a^4+\cdots\\ =(a+a^2+a^3+a^4+\cdots)+a(a+a^2+a^2+a^3+a^3+a^3+\cdots),$$

hence for $|a|<1$,

$$S=\frac a{1-a}+aS.$$