I meet a problem with calculating $$ \frac{e^{-|\alpha|^2}}{\sum A_k^2}\left|\sum_{k=0}^{\infty} A_k \frac{(-\alpha)^k}{\sqrt{k!}}\right|^2\,. $$ Only I know is the constrain that $\sum_{k=0}^{\infty} A_k \frac{\alpha^k}{\sqrt{k!}}=0$.
How to get the result, or the upper and lower bound?
(so sorry that I missed the square roof of $k!$)
I believe the purpose of that constraint is to just prove that the series converges, as if that sequence converges, that alternating sequence does too.
Using @Dmitry's hint, you can get a decent upper bound with Cauchy-Schwartz:
$|U*V|^2 \le U^2 * V^2 $, which with $\sum U_k * V_k \le \sum U_k \sum V_k$ implies that $\sum |U * V|^2 = |\sum U_k|^2|\sum V_k|^2$ $$ \frac{e^{-|\alpha|^2}}{\sum A_k^2}\left|\sum_{k=0}^{\infty} A_k \frac{(-\alpha)^k}{k!}\right|^2 \le \frac{e^{-|\alpha|^2}}{\sum A_k^2}\left|\sum_{k=0}^{\infty} A_k \right|^2 \left| \sum_{k=0}^{\infty}\frac{(-\alpha)^k}{k!}\right|^2 $$
$$ \le \frac{e^{-|\alpha|^2}}{\sum A_k^2}(\sum_{k=0}^{\infty} A_k^2) \left| \sum_{k=0}^{\infty}\frac{(-\alpha)^k}{k!}\right|^2 $$
$$ = e^{-|\alpha|^2} \left|\sum_{k=0}^{\infty}\frac{(-\alpha)^k}{k!}\right|^2 $$
$$ = e^{-|\alpha|^2} (e^{-\alpha})^2 $$
$$ = e^{-|\alpha|^2 - 2\alpha} $$
The simplest lower bound would be zero. This could occur for $\alpha = 0$ or specific $A$'s, whether it makes $lim_{n->\infty} \sum A_k^2 = \infty$ or the alternating sequence sum is pushed to zero somehow.