Limit $\lim_{x\rightarrow +\infty}\sqrt{x}e^{-x}\left(\sum_{k=1}^{\infty}\frac{x^{k}}{k!\sqrt{k}}\right)$

Question

$$\lim_{x\rightarrow +\infty}\sqrt{x}e^{-x}\left(\sum_{k=1}^{\infty}\frac{x^{k}}{k!\sqrt{k}}\right)$$ Any hint will be appreciated.

Note: There is a related question on MathOverflow: Asymptotic expansion of $\sum\limits_{n=1}^{\infty} \frac{x^{2n+1}}{n!{\sqrt{n}}}$. The MO question references this question and it also links to some other posts containing similar expressions.

Answer 1

(I have overwritten my previous incorrect answer.)

For any positive constant $c > 1$ we have: \begin{align*}S(cx) = e^{-x}\sum\limits_{k \ge cx} \frac{x^k}{k!} &= e^{-x}\frac{x^{cx}}{(cx)!}\left(1 + \frac{x}{(cx+1)} + \frac{x^2}{(cx+1)(cx+2)} + \cdots\right) \\& \le e^{-x}\frac{x^{cx}}{(cx)!}\sum\limits_{k=0}^{\infty}\frac{1}{c^k} < \frac{c}{c-1}e^{-x}x^{cx}\frac{e^{cx}}{(cx)^{cx}} = \frac{ce^{-((c\log c) -c +1)x}}{c-1}\end{align*}

where, we used the inequality, $x! > e^{-x}x^{x}$ for sufficiently large $x$. Also note that $(c\log c) - c + 1 > 0$ for all $c > 0$

Again, for sufficiently large $x$: \begin{align*}T(x/c) = e^{-x}\sum\limits_{0 \le k < x/c} \frac{x^k}{k!} = e^{-x}\sum\limits_{0 \le k < x/c} \frac{c^k(x/c)^k}{k!} & \le e^{-x}\frac{(x/c)^{x/c}}{(x/c)!}\sum\limits_{0 \le k < x/c} c^k \\& < e^{-x}e^{x/c}\frac{xc^{x/c}}{c} = \frac{x}{c}e^{-\left(1 - \frac{1}{c} - \frac{\log c}{c}\right)x}\end{align*} where, we note that $(1 - \frac{1}{c} - \frac{\log c}{c}) > 0$.

Hence, we have for any $c>1$, $$\displaystyle e^{-x}\sum\limits_{x/c \le k \le cx} \frac{x^k}{k!} = 1 - S(cx) - T(x/c) \quad \underbrace{\longrightarrow}_{ x \to +\infty} \quad 1 \tag{1}$$

Now, we have the simple estimates, $$\frac{e^{-x}}{\sqrt{cx}}\sum\limits_{x/c \le k \le cx} \frac{x^k}{k!} < e^{-x}\sum\limits_{x/c \le k \le cx} \frac{x^k}{k!\sqrt{k}} < \frac{e^{-x}}{\sqrt{x/c}}\sum\limits_{x/c \le k \le cx} \frac{x^k}{k!} \tag{2}$$

and $$e^{-x}\sum\limits_{k \not\in [x/c,cx]} \frac{x^k}{k!\sqrt{k}} < T(x/c) + S(cx) \tag{3}$$ where, both the terms in the upper bound decays exponentially.

Hence, from $(1),(2)$ and $(3)$ we have that the required limit: $$\displaystyle \lim\limits_{x \to \infty} \sqrt{x}e^{-x}\sum\limits_{k=1}^{\infty} \frac{x^k}{k!\sqrt{k}} \in \left(\frac{1}{\sqrt{c}},\sqrt{c}\right)$$ for arbitrary $c > 1$, i.e., the required limit is $1$ (as pointed out by user tired in the comments).

I suppose for any function $\alpha(x)$ with atmost polynomial growth and continuous at $x = 1$ we can slightly modify the above argument and show, $$\lim\limits_{x \to \infty} e^{-x}\sum\limits_{k=1}^{\infty} \alpha\left(\frac{k}{x}\right)\frac{x^k}{k!} = \alpha(1)$$