I've been trying to find an asymptotic expansion of the following series


$$C(x) = \sum\limits\_{n=1}^{\infty} \frac{x^{2n+1}}{n!{\sqrt{n}} }$$


and


$$L(x) = \sum\limits\_{n=1}^{\infty} \frac{x^{2n+1}}{n(n!{\sqrt{n}}) }$$


around $+\infty$, in the from


$$\exp(x^2)\Big(1+\frac{a\_1}{x}+\frac{a\_2}{x^2} + .. +\frac{a\_k}{x^k}\Big) + O\Big(\frac{\exp(x^2)}{x^{k+1}}\Big)$$


where $x$ is a positive real number. As far as I progressed, I obtained only


$$C(x) = \exp(x^2) + \frac{\exp(x^2)}{x} + O\Big(\frac{\exp(x^2)}{x}\Big).$$


I tried to use ideas from <https://math.stackexchange.com/questions/484367/upper-bound-for-an-infinite-series-with-a-square-root?rq=1>, <https://math.stackexchange.com/questions/115410/whats-the-sum-of-sum-limits-k-1-infty-fractkkk>, <https://math.stackexchange.com/questions/378024/infinite-series-involving-sqrtn?noredirect=1&lq=1>, but I was unable to make them work in my case. 


Any suggestions would be greatly appreciated!


(If someone has a solid culture in this kind of things, is there are any specific names for $C(x) $ and $L(x) $ ?).


PS:


This question was asked on the math.SE but was closed as duplicate of <https://math.stackexchange.com/questions/2117742/lim-x-rightarrow-infty-sqrtxe-x-left-sum-k%ef%bc%9d1-infty-fracxk/2123100#2123100>. However, the latter question provides only the first term of the asymptotic expansion and does not address sufficiently the problem considered here.



Asymptotic expansion of $\sum\limits_{n=1}^{\infty} \frac{x^{2n+1}}{n!{\sqrt{n}} }$
I sketch the arguments for $C(x)$, the arguments for $L(x)$ are essentially the same.


The specific form of the sum suggests probabilistic arguments. 
Let $X\_x$ be a $\mathrm{Poiss}(x^2)$-distributed random variable and note that
$$C(x)\,e^{-x^2}=\mathbb{E}\frac{x}{\sqrt{X\_x}}\,1\_{\{X\_x\geq 1\}}$$ 
It is known that $\frac{X\_x -x^2}{x}\longrightarrow N(0,1)$ (standard normal) in distribution as $x\longrightarrow \infty$ and that
that all moments $\mathbb{E}\left(\frac{X\_x-x^2}{x}\right)^k$ of $X\_x$ converge to the corresponding moments of $N(0,1)$, so that (for large $x$) $X\_x$ is concentrated around $x^2$, with deviations of order $x$.


To use that information split $\{X\_x\geq 1\}$ on the rhs into (say) the parts $1\leq X\_x <\tfrac{1}{2}x^2$, $X\_x-x^2 >\tfrac{1}{2} x^{2}$ and
$|X\_x-x^2| \le \tfrac{1}{2}x^{2}$.


By routine arguments the integrals over the first two parts are asymptotically exponentially small.
For the remaining part write
\begin{align\*}
 \mathbb{E}\frac{x}{\sqrt{X\_x}}\,1\_{\{|X\_x-x^2|\leq \tfrac{1}{2}x^2\}} 
 &=\mathbb{E}\frac{x}{\sqrt{x^2+(X\_x-x^2)}}\,1\_{\{|X\_x-x^2|\leq \tfrac{1}{2} x^2\}}\\
 &=\mathbb{E}\frac{1}{\sqrt{1 +\frac{1}{x^2}(X\_x-x^2)}}\,1\_{\{|X\_x-x^2|\leq \tfrac{1}{2} x^2\}}\\
 &=\mathbb{E}\sum\_{k=0}^\infty {-\frac{1}{2} \choose k} \frac{1}{x^{2k}}(X\_x-x^2)^k
 \,1\_{\{|X\_x-x^2|\leq \tfrac{1}{2}x^2\}}\\
%C(x)\,e^{-x^2}&= 1+\frac{3}{8}x^{-2} + \frac{65}{128}x^{-4} + \frac{1225}{1024}x^{-6} + \frac{1619583}{425984}x^{-8}+\mathcal{O}(x^{-10})\\
%L(x)\,x^2\,e^{-x^2}&= 1+\frac{15}{8}x^{-2} + \frac{665}{128}x^{-4}+\frac{19845}{1024}x^{-6}+\frac{37475823}{425984}x^{-8}+\mathcal{O}(x^{-10})
\end{align\*}
Clearly the series may be integrated termwise, and completing the tails changes it only by asymptotically exponentially small terms. For $k\geq 2$ the central moment
$c\_k(x):=\mathbb{E}\left(X\_x-x^2\right)^k$ is a polynomial in $x^2$ of degree $\lfloor k/2\rfloor$.
Thus (after regrouping of terms) the formal series
$$\sum\_{k=0}^\infty x^{-2k}{-\tfrac{1}{2} \choose k} c\_k(x)$$
gives a full asymptotic expansion of $C(x)e^{-x^2}$. Evaluating the first eight terms gives
$$C(x)\,e^{-x^2}= 1+\frac{3}{8}x^{-2} + \frac{65}{128}x^{-4} + \frac{1225}{1024}x^{-6} + \frac{131691}{32768}x^{-8}+\mathcal{O}(x^{-10})$$


**EDIT**: I corrected the coefficient of $x^{-8}$. Thanks to Johannes Trost
for pointing out that I miscalculated. Similarly the coefficient of $x^{-8}$
in the asymptotic series for $L(x)$ given in the comment must be corrected.


