What is the range of convergence of $\sum_{k=0}^\infty (k \cdot x \exp(-x))^k\cdot {1 \over k!} \cdot {1\over k+1}$? I have the following series as an expression which occurs as a limit of a quotient of polynomials in $e$ and $x$ which I've expanded by polynomial long division into a series:
$$f(x) = \sum_{k=0}^\infty \Big({k \cdot x \over e^x}\Big)^k\cdot {1 \over k!} \cdot 
{1\over k+1}$$
I'm interested in the convergence/divergence of this power series for the argument in the range $ x \gt 0$.

This is what I have done so far:
Let
$$ c_k(x)=\Big({k \cdot x \over e^x}\Big)^k\cdot {1 \over k!}  $$
and let by the Stirling-approximation
$$\Big({k \over e}\Big)^k\cdot {1 \over k!} \approx {1\over \sqrt{2 \pi }}{1\over \sqrt{k}} $$

Then for the k'th term I get      


*

*with $x=1$ the approximation
$$ c_k(1)\cdot {1\over k+1} \approx {1\over \sqrt{2 \pi }}{1\over k^{1.5}} $$
I think this is sufficient to show convergence (compared for instance with the Dirichlet-series zeta(1.5)).        

*If $x \gt 1$ the terms $c_k(x)$ decrease and thus I think the series is obviously even more convergent, so I assume that we have convergence for $x \ge 1$.         

However, for $0<x<1$ it is not so obvious at the moment.          


Q: Is there a (possibly similar simple) argument for the convergence in the remaining range $0<x<1$ ?


Substitute $y = xe^{-x}$. Then, the series $\displaystyle\sum_{k=0}^\infty (k \cdot y)^k\cdot {1 \over k!} \cdot {1\over k+1}$ converges for all $|y| < R$ for some radius of convergence $R$. (This is a property of power series)
As you have shown, the series converges for $x = 1$, i.e. $y = \dfrac{1}{e}$. So, we know that $R \ge \dfrac{1}{e}$. 
The function $f(x) = xe^{-x}$ has derivative $f'(x) = (1-x)e^{-x}$. So, $f'(x) > 0$ for $0 < x < 1$, and $f'(x) < 0$ for $x > 1$. Therefore, for all $x \ge 0$, we have $f(x) \le f(1)$, i.e. $y = xe^{-x} \le \dfrac{1}{e} \le R$. 
Therefore, the series converges for all $x \ge 0$. 
