# 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!

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 $b\leq1$ is pretty easy. But $b>1$ is killing me :( – wircho May 8 '12 at 4:16 Got it. I will post my own answer! – wircho May 8 '12 at 4:42

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 $a+1>1$.

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 This bound only works for $a\geq 1$. The sum is not bounded by $C e^b$ for $0 For convenience, we consider the sum starting at$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\$.

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