# Limit of $\frac{a^{n+1}(n+1)!^b}{\sum_{k=0}^n a^kk!^b}$ when $n\rightarrow\infty$

I want to prove that $$\lim_{n\rightarrow\infty}\frac{a^{n+1}(n+1)!^b}{\sum_{k=0}^n a^kk!^b}<\infty$$ for $a,b>0$.

This is the last step of a bigger problem. I believe it would suffice to use good enough upper and lower bounds for the factorials, but I don't know such bounds. Any help would be greatly appreciated!

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Without further conditions on $a$ and $b$, this cannot be true. For example, let $a=b=1$, and let $n \ge 1$. Then the numerator is $(n+1)!$. The denominator is $\le n(n-1)!+n!$, that is, $\le 2(n!)$, so the ratio is $\ge (n+1)/2$.
I originally thought $0 < a < 1$ would suffice, but I think even this is wrong. If $a = 1/2$, and $b = 1$, then the limit is still $\infty$, for the same reason here (with a slightly modified bound). –  Nicholas Stull May 8 '12 at 3:22
@NicholasStull: Yes, $b=1$ is a problem in any case. –  André Nicolas May 8 '12 at 3:43
I am beginning to wonder if this statement is actually false for $b > 0$, if $b$ is constant (with respect to $n$). I suspect we must require $b\to 0$ as $n\to\infty$. But even that may not be sufficient. –  Nicholas Stull May 8 '12 at 3:56
@wircho, thanks for the reference to the other problem. I will look into this, but I suspect it may be possible to prove the limit in this question exists if $b = \frac{1}{n^2}$. –  Nicholas Stull May 8 '12 at 4:08