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](https://mathoverflow.net/questions/161068/has-anyone-seen-this-series), then an upper bound can be obtained, which is $1$. This problem comes from estimation of the behavior of the integral in [this post](https://mathoverflow.net/questions/161659/how-to-integrate-an-exponential-function-of-an-exponential-function).


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



Looking for a limit related to the series in a previous post
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. 


