Leibniz test and $\sum\frac{(-1)^k k^2}{\sqrt{k!}}$ I would like to check if the following series $\sum\frac{(-1)^k k^2}{\sqrt{k!}}$ is convergent.
My original idea was to employ Leibniz's criterion, the one that says: An alternating series $(-1)^k a_k$ such that $a_k$ is a non-increasing sequence that satisfies $\lim a_k=0$ then the series converges, however I'm having difficults when showing that my $a_k$ is non-increasing through induction.
I would appreciate some help.
 A: [Note] : $a_{n}>0$ is non-increasing if and only if $\displaystyle\frac{a_{n+1}}{a_{n}}\leq1$.
$$
\frac{a_{n+1}}{a_{n}}=\frac{\frac{(n+1)^{2}}{\sqrt{(n+1)!}}}{\frac{n^{2}}{\sqrt{n!}}}=\frac{(n+1)^{2}}{n^{2}}\frac{\sqrt{n!}}{\sqrt{(n+1)!}}=\frac{(n+1)^{2}}{n^{2}}\frac{1}{\sqrt{n+1}}
$$
$$
\lim_{n\to\infty}\frac{(n+1)^{2}}{n^{2}}\frac{1}{\sqrt{n+1}}=1\cdot0=0<1
$$
Clearly non-increasing.
A: $$a_n=\frac{k^2}{\sqrt{k!}}\implies \log(a_n)=2\log(k)-\frac 12\log(k!)$$ Using Stirling approximation
$$\log(a_n)=\frac 12(1-\log(k))k+\cdots$$
$$a_n=e^{\log(a_n)}=e^{\frac{1}{2} k (1-\log (k))}~~~\to ~~~0$$
A: It is enough to show that the terms $a_k$ are non-increasing for all $k$ larger than a fixed finite number (Can you see why?). For $k \ge 3$,
$$
\frac{{\frac{{(k + 1)^2 }}{{\sqrt {(k + 1)!} }}}}{{\frac{{k^2 }}{{\sqrt {k!} }}}} = \left( {1 + \frac{1}{k}} \right)^2 \frac{1}{{\sqrt {k + 1} }} < \left( {1 + \frac{1}{3}} \right)^2 \frac{1}{{\sqrt {3 + 1} }} = \frac{8}{9} < 1
$$
Thus
$$
\frac{{(k + 1)^2 }}{{\sqrt {(k + 1)!} }} < \frac{{k^2 }}{{\sqrt {k!} }}
$$
for $k\geq 3$, i.e., the terms are decresing starting from the third term.
