Upper bound for the series $\sum_{n\geq 1}\frac{1}{(n+1)^{a+1}}\sum_{k=0}^n b^k\left(\frac{(n-k)!}{n!}\right)^a$ I want to show that the series
$$\sum_{n\geq 1}\frac{1}{(n+1)^{a+1}}\sum_{k=0}^n b^k\left(\frac{(n-k)!}{n!}\right)^a$$
converges for $a,b>0$. I have tried this so much that the smallest hint will probably suffice. I asked a question before which would have been enough but it is not true. Right now I am really stuck and frustrated. Any help would be greatly appreciated!
 A: It is enough to show that the sum for $n\geq0$ converges. Changing sums and manipulating I get:
$$\sum_{k\geq0}\frac{b^k}{k!}\sum_{n\geq k}\frac{1}{(n+1)^{a+1}\left(\begin{array}{c}n\\ k\end{array}\right)}$$
$$\leq\sum_{k\geq0}\frac{b^k}{k!}\sum_{n\geq k}\frac{1}{(n+1)^{a+1}}$$
$$\leq\sum_{k\geq0}\frac{b^k}{k!}\sum_{n\geq 0}\frac{1}{(n+1)^{a+1}}$$
$$=\sum_{k\geq0}\frac{b^k}{k!}C=Ce^{b}$$
where $C$ is a constant $<\infty$ because $a+1>1$.
A: For convenience, we consider the sum starting at $n=0$. 
Then 
$$\begin{eqnarray*}
\sum_{n=0}^\infty \frac{1}{(n+1)^{a+1}}\sum_{k=0}^n b^k\left(\frac{(n-k)!}{n!}\right)^a
&=&  \sum_{k=0}^\infty \frac{b^k}{(k!)^a} 
\sum_{n=k}^\infty \frac{1}{(n+1)^{a+1}} \frac{1}{{n\choose k}^{a}} \\
&\leq& \sum_{k=0}^\infty \frac{b^k}{(k!)^a}
\sum_{n=k}^\infty \frac{1}{(n+1)^{a+1}} \\
&\leq& \zeta(a+1) \sum_{k=0}^\infty \frac{b^k}{(k!)^a}. \\
\end{eqnarray*}$$
We have used the fact that $1/{n\choose k}^a \leq 1$ for $a>0$. 
Notice that $k!^a \geq k!$ only if $a\geq 1$. 
Thus, the sum is bounded by $\zeta(a+1) \sum_{k=0}^\infty b^k/k! = \zeta(a+1)e^b$ only if $a\geq 1$.
The sum converges if $\sum_{k=0}^\infty b^k/(k!)^a$ converges.
But the ratio of successive terms goes like
$b/k^a$, and so vanishes in the limit since $a>0$.
Thus, the series converges.
Notice the convergence of $\sum_{k=0}^\infty b^k/(k!)^a$ can be very slow.
Let $a=1/10$ and $b=10$.
It is not until we reach $k=10^{10}$ that the ratio of successive terms is less than $1$.
