Can I know the value of an infinite serie?

Question

$$\sum\limits_{n=0}^{\infty}\frac{n}{e^n}$$

I have found through a software that the value is $\dfrac{e}{(e-1)^2}$.

I've been trying to do it manually but I am getting $\dfrac{\infty}{\infty}$, since I have in the

Answer 1

Note that $f(x)=\sum\limits_{k=0}^\infty x^n=\frac{1}{1-x}$ for $|x|<1$. Differentiating this, we have

$$f'(x)=\sum_{k=0}^\infty nx^{n-1}=\frac1{(1-x)^2}$$

$$\implies xf'(x)=\sum_{k=0}^\infty nx^{n}=\frac x{(1-x)^2}$$

For $|x|<1$. Since $0<e^{-1}<1$, we can substitute it into the expression to recieve

$${e}^{-1}f'(e^{-1})=\sum_{k=0}^\infty ne^{-n}=\frac {e^{-1}}{(1-e^{-1})^2}=\frac{e}{(e-1)^2}$$

Answer 2

Consider, instead, the series: $$\begin{align} \sum_{n=0}^\infty nx^n &= x\sum_{n=0}^\infty nx^{n-1} \\ &= x\sum_{n=0}^\infty\frac{d}{dx}x^n \\ &= x\frac{d}{dx}\sum_{n=0}^\infty x^n \\ &= x\frac{d}{dx}\left(\frac{1}{1-x}\right) \\ &= x\left(\frac{1}{(1-x)^2}\right) \\ &= \frac{x}{(1-x)^2} \end{align}$$

This converges whenever $|x|< 1$. By making a smart substitution, you have your result.

Answer 3

Note $$S_n= \frac{1}{e}+\frac{2}{e^2}+\cdots+\frac{n}{e^n}$$ It is obtained $S_n$ by calculating the difference $$S_n-\frac{1}{e}S_n$$and then get to the limit.