TITLE: Determine for which values the series $\sum_{k=1}^\infty \frac{b^{k^{2}}}{k!}$ is convergent.
QUESTION [5 upvotes]: So I am trying to figure out for which positive real numbers b this series

$$\sum_{k=1}^\infty \frac{b^{k^{2}}}{k!}
$$ 

is convergent.
Since it looks pretty hard to compare it with other series, I decided to use the ratio test.
$$\lim_{k \to \infty} \left|\frac{b^{(k+1)^{2}}}{(k+1)!}\cdot{\frac{k!}{b^{k^{2}}}}\right| $$ 
$$=\lim_{k \to \infty} \left|\frac{b^{2k+1}}{k+1}\right| < 1 $$
But I am stuck here, this doesnt look too useful. Looking for some help.

REPLY [4 votes]: From your attempt one gets that, for $|b|\le1$, as $k \to \infty$,
$$
\left| \frac{u_{k+1}}{u_k}\right|=\left|\frac{b^{2k+1}}{k+1}\right|\le \frac{|b|^{2k+1}}{k+1} \le \frac1{k+1} \to 0
$$ Then, by the ratio test the series is convergent for $|b|\le 1$.
One may observe that, for $|b|>1$, 
$$
\lim_{k \to \infty}\left| \frac{u_{k+1}}{u_k}\right|= \lim_{k \to \infty}\frac{|b|^{2k+1}}{k+1} =\infty
$$ and the series is divergent for $|b|>1$.