# Find the sum of the infinite series -ne^(1-n) from n=1 to oo?

Mar 5, 2018

$- {e}^{2} / {\left(e - 1\right)}^{2}$

#### Explanation:

The simplest solution I can think of starts with the infinite geometric series

$S \left(a\right) = {\sum}_{n = 1}^{\infty} {e}^{- a n}$

which converges for $a > 0$ to the well known sum

$S \left(a\right) = {e}^{- a} / \left\{1 - {e}^{- a}\right\} = \frac{1}{{e}^{a} - 1}$

Differentiating both sides with respect to $a$ leads to

$\frac{d}{\mathrm{da}} \left({\sum}_{n = 1}^{\infty} {e}^{- a n}\right) = \frac{d}{\mathrm{da}} \left(\frac{1}{{e}^{a} - 1}\right) \implies$
$- {\sum}_{n = 1}^{\infty} n {e}^{- a n} = - {e}^{a} / {\left({e}^{a} - 1\right)}^{2}$

Substituting $a = 1$ gives us

$- {\sum}_{n = 1}^{\infty} n {e}^{- n} = - \frac{e}{e - 1} ^ 2 \implies$
$- {\sum}_{n = 1}^{\infty} n {e}^{1 - n} = - {e}^{2} / {\left(e - 1\right)}^{2}$

Alternative

We can evaluate the sum by a purely algebraic approach as well:

$- S = 1 + \frac{2}{e} + \frac{3}{e} ^ 2 + \frac{4}{e} ^ 3 + \ldots$

$- \frac{1}{e} S = \frac{1}{e} + \frac{2}{e} ^ 2 + \frac{3}{e} ^ 3 + \ldots$

Subtracting, we get

$- \left(1 - \frac{1}{e}\right) S = 1 + \frac{1}{e} + \frac{1}{e} ^ 2 + \frac{1}{e} ^ 3 + \ldots = \frac{1}{1 - \frac{1}{e}}$