power series of function I am studying on summation theory on power series of functions.
My question is to find the sum of power series  $$e^{-n} \sum_{k=0}^{\infty} \frac{n^k}{k!}\frac{k}{k+1}.$$ I tried apply differentiation and integration but could not find anything useful. Thanks in advance from now.
 A: You are starting with
$$ e^{-x} \sum_{k = 1}^\infty \frac{kx^k}{(k+1)!} = e^{-x}F(x),$$
which is almost very easy. Let's ignore the $e^{-x}$ piece, because it's not interesting. So we just consider $F(x)$.
Notice that $\dfrac{F(x)}{x} = \displaystyle\sum_{k = 1}^\infty \frac{kx^{k-1}}{(k+1)!}.$ If we integrate this, we see that
$$\begin{align} \int_0^x \frac{F(t)}{t} dt &= C + \sum_{k = 1}^\infty \frac{x^k}{(k+1)!} = C + \frac{1}{x}\sum_{k = 2}^\infty \frac{x^k}{k!} \\
&= C - \frac{1}{x} - 1 + \frac{1}{x} \sum_{k = 0}^\infty \frac{x^k}{k!} \\
&= \frac{e^x}{x} - \frac{1}{x} - 1 + C.
\end{align}$$
We are interested in $F(x)$, which we recover from the fundamental theorem of calculus.
$$\begin{align}
\frac{d}{dx} \int_0^x \frac{F(t)}{t} dt &= \frac{F(x)}{x} \\
&= \frac{d}{dx} \left( \frac{e^x}{x} - \frac{1}{x} - 1 + C \right) \\
&= \frac{e^x}{x} - \frac{e^x}{x^2} + \frac{1}{x^2}
\end{align}$$
Thus
$$ F(x) = e^x - \frac{e^x}{x} + \frac{1}{x},$$
so that 
$$e^{-x}F(x) = 1 - \frac{1}{x} + \frac{e^{-x}}{x},$$
which is your answer. $\diamondsuit$
