Upper bound for the series $\sum_{n\geq 1}\frac{1}{(n+1)^{a+1}}\sum_{k=0}^n b^k\left(\frac{(n-k)!}{n!}\right)^a$

Question

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!

Answer 1

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 $a+1>1$.

Answer 2

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 $1$.