Looking for a limit related to the series in a previous post Can any one show that the following limit?
$$
\lim_{z\rightarrow \infty}  \sqrt{z} \: e^{-z}\sum_{k=1}^\infty \frac{z^k}{k! \sqrt{k}} \quad \stackrel{?}{=} \quad\sqrt{2}-1.
$$
If one uses the estimate based on my previous post, then an upper bound can be obtained, which is $1$. This problem comes from estimation of the behavior of the integral in this post.
EDIT：
Lucia and Lucian gave some hints and numerics. But I still don't have a clue to prove this limit rigorously. Does any one have any clue?
Thanks a lot! Anand
 A: Assume below that $z$ is positive.  Note that, using $\sqrt{z}-\sqrt{k} = (z-k)/(\sqrt{z}+\sqrt{k})$, 
$$ 
\Big| \frac{e^z}{\sqrt{z}} - \sum_{k=1}^{\infty} \frac{z^k}{k!\sqrt{k}} \Big| 
\le \frac{1}{\sqrt{z}} + \Big|\sum_{k=1}^{\infty} \frac{z^k}{k!} \frac{\sqrt{z}-\sqrt{k}}{\sqrt{kz}} \Big| \le \frac{1}{\sqrt{z}}+ \sum_{k=1}^{\infty} \frac{z^k}{k!} \frac{|z-k|}{\sqrt{k} z}. \tag{1} 
$$
Now by Cauchy-Schwarz 
$$ 
\sum_{k=1}^{\infty} \frac{z^k}{k!}\frac{|z-k|}{\sqrt{k}} \le \Big( \sum_{k=1}^{\infty} \frac{z^k}{k!} \frac{1}{k} \Big)^{\frac 12} \Big( \sum_{k=1}^{\infty} \frac{z^k}{k!} (z-k)^2 \Big)^{\frac 12},
$$ 
and note that 
$$ 
\sum_{k=1}^{\infty} \frac{z^k}{k!}\frac{1}{k} \le 2\sum_{k=1}^{\infty} \frac{z^{k}}{k!} \frac{1}{(k+1)} \le \frac{2}{z} e^z, 
$$ 
and 
$$ 
\sum_{k=1}^{\infty} \frac{z^k}{k!} (z-k)^2 \le \sum_{k=0}^{\infty} \frac {z^k}{k!}(z-k)^2 = ze^z.
$$ 
Using these bounds in (1), we get 
$$ 
\Big| \frac{e^z}{\sqrt{z}} - \sum_{k=1}^{\infty} \frac{z^k}{k!\sqrt{k}} \Big| 
\le \frac{1}{\sqrt{z}} + \sqrt{2} \frac{e^z}{z}.
$$ 
It readily follows that the limit in the question equals $1$.  One can be more precise, but this is simple and good enough. 
A: Let $X$ be a random variable with Poisson($\lambda$) distribution. Let $Y$ be the random variable where $Y=(\lambda/X)^{1/2}$ if $X\neq 0$ and $Y=0$ otherwise.  We are asked take the limit of $\mathbb{E}Y$ as $\lambda\to\infty$.
The variable $X$ has mean $\lambda$ and standard deviation $\sqrt\lambda$.  By Chebyshev's inequality, for every $T>0$ the probability that $|X - \lambda| \leq T\sqrt\lambda$ is at least $1-T^{-2}$.  For $X$ in that range, $Y$ is $1+O_T(\lambda^{-1/2})$.  Since $Y\leq\sqrt\lambda$, it follows that
$$ \left|\mathbb{E}Y-1\right| = O(T\lambda^{-1/2} + T^{-2}\lambda^{1/2})\,.$$
Now taking $T=\lambda^{1/3}$ gives
$$ \left|\mathbb{E}Y-1\right| = O(\lambda^{-1/6})\,.$$
In fact, as $\lambda\to\infty$, $X'=\frac{X-\lambda}{\sqrt{\lambda}}$ converges in distribution to a standard normal variable, which should give the correct rate of convergence.
