$\sum\limits_{n=1}^{\infty} \frac{1}{ne^{n}}$ converges

Question

Here, The WolframAlpha calculates this series :

$\sum\limits_{n=1}^{\infty} \frac{1}{ne^{n}} = 1 - \ln (e -1)$

But they do not explain how they do this calculation. Does anyone know how to do it?

Thanks and regards.

Answer 1

hint

Consider the well-known result

$$\sum_1^\infty x^{n-1}=\frac{1}{1-x}$$

for $$|x|<1$$

if we integrate, it gives

$$\sum_{n=1}^\infty\frac{x^{n}}{n}=-\ln(1-x)$$

Now, replace $x$ by $\frac 1e$ and observe that

$$\ln(1-\frac 1e)=\ln(\frac{e-1}{e})$$ $$=\ln(e-1)-\ln(e)=\ln(e-1)-1$$

Answer 2

We know how to compute sum of geometric series. For $x\in[0,1)$ we have $\sum_{n=1}^\infty x^{n-1}=\frac{1}{1-x}$. Now there is a theorem which says we can integrate power series term by term. So for given $t\in[0,1)$ we can integrate both sides in the interval $[0,t]$ and get $\sum_{n=1}^\infty\frac{t^n}{n}=-ln(1-t)$. Now put $t=\frac{1}{e}$ and you will get the result you need.

Answer 3

Polylogarithm function: $Li_s(z)=\sum\limits_{n=1}^\infty \frac{z^n}{n^s}$

in this case $s=1$ and using that $Li_1(z)=-\ln(1-z)$

as $z=\frac{1}{e}$ we get the result: $\sum\limits_{n=1}^{\infty} \frac{1}{ne^{n}}=-\ln(1-\frac{1}{e})=\ln(\frac{e}{e-1})=1-\ln(e-1)$