Upper bound for $S (x) = e^{-x} \sum_{k=0}^{\infty} \frac{x^k}{k!} \sqrt{k+1}$

Question

I've been trying to find a tight upper bound for the series

$$S (x) = e^{-x} \sum_{k=0}^{\infty} \frac{x^k}{k!} \sqrt{k+1}$$

So far, I've managed to get a reasonable bound for small values of $x$ by using the inequality $\sqrt{k+1} \leq \sqrt{\frac{k^{2}}{4} + k + 1} = \frac{k}{2} + 1 ~\forall~k \geq 0$, but it becomes very loose when $x$ is large. I've also tried taking a Taylor series approximation to $\sqrt{k+1}$, but this leads to a complicated infinite sum of weighted Bell polynomials which, as far as I'm aware, doesn't have a closed form. Any suggestions would be greatly appreciated!

Answer 1

Upper and Lower Bounds

Note that $$ e^{-x}\sum_{k=0}^\infty\frac{x^k}{k!}=1\tag{1} $$ and that $$ e^{-x}\sum_{k=0}^\infty(k+1)\frac{x^k}{k!}=x+1\tag{2} $$ Since $\sqrt{x}$ is concave, Jensen's Inequality gives $$ e^{-x}\sum_{k=0}^\infty\sqrt{k+1}\frac{x^k}{k!}\le\sqrt{x+1}\tag{3} $$ Also, $$ e^{-x}\sum_{k=0}^\infty\frac1{k+1}\frac{x^k}{k!}=\frac{1-e^{-x}}{x}\tag{4} $$ Since $1/\sqrt{x}$ is convex, Jensen's Inequality gives $$ \begin{align} e^{-x}\sum_{k=0}^\infty\sqrt{k+1}\frac{x^k}{k!} &\ge\sqrt{\frac{x}{1-e^{-x}}}\\ &\ge\sqrt{x}\tag{5} \end{align} $$ Therefore, we get the bounds $$ \sqrt{x}\le e^{-x}\sum_{k=0}^\infty\sqrt{k+1}\frac{x^k}{k!}\le\sqrt{x+1}\tag{6} $$

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Asymptotic Expansion

Using Stirling's Expansion and the Binomial Theorem, we get $$ \begin{align} \frac1{4^n}\binom{2n}{n} &=\frac1{\sqrt{\pi n}} \left(1-\frac1{8n}+\frac1{128n^2}+\frac5{1024n^3}-\frac{21}{32768n^4}+\dots\right)\\ &=\frac1{\sqrt{\pi(n+1)}} \left(1+\frac3{8n}-\frac{23}{128n^2}+\frac{89}{1024n^3}-\frac{1509}{32768n^4}+\dots\right)\tag{7} \end{align} $$ and therefore, $$ \begin{align} \frac{\sqrt{n+1}}{n!} &=\frac{4^n}{\sqrt{\pi}}\frac{n!}{(2n)!}\left(1+\frac3{8n}-\frac{23}{128n^2}+\frac{89}{1024n^3}-\frac{1509}{32768n^4}+\dots\right)\\ &=\frac{2^n}{\sqrt{\pi}}\frac1{(2n-1)!!}\left(1+\frac3{8n}-\frac{23}{128n^2}+\frac{89}{1024n^3}-\frac{1509}{32768n^4}+\dots\right)\\ &=\frac{2^n}{\sqrt{\pi}}\small\left(\frac1{(2n{-}1)!!}+\frac{3/4}{(2n{+}1)!!}+\frac{1/32}{(2n{+}3)!!}+\frac{9/128}{(2n{+}5)!!}+\frac{491/2048}{(2n{+}7)!!}+\dots\right)\tag{8} \end{align} $$ Note that $$ \begin{align} \int_x^\infty e^{-t^2/2}\,\mathrm{d}t &=\frac1x\int_x^\infty\frac{x}{t}e^{-t^2/2}\,\mathrm{d}t^2/2\\ &\le\frac1x\int_x^\infty e^{-t^2/2}\,\mathrm{d}t^2/2\\ &=\frac1xe^{-x^2/2}\tag{9} \end{align} $$ therefore, since both the following sum and integral satisfy $f'=1+xf$ and agree at $x=0$, $$ \begin{align} \sum_{k=0}^\infty\frac{x^{2k+1}}{(2k+1)!!} &=e^{x^2/2}\int_0^xe^{-t^2/2}\,\mathrm{d}t\\ &=\sqrt{\frac\pi2}\ e^{x^2/2}+O\left(\frac1x\right)\\ \frac1{\sqrt{2x}}\sum_{k=0}^\infty\frac{(2x)^{k+1}}{(2k+1)!!} &=\sqrt{\frac\pi2}e^x+O\left(\frac1{\sqrt{x}}\right)\\ e^{-x}\sum_{k=0}^\infty\frac{(2x)^{k+1}}{(2k+1)!!} &=\sqrt{\pi x}+O\left(e^{-x}\right)\tag{10} \end{align} $$ Multiplying $(8)$ by $e^{-x}x^n$, summing, and applying $(10)$ yields the asymptotic expansion that Raymond Manzoni got: $$ \begin{align} e^{-x}\sum_{n=1}^\infty\frac{\sqrt{n+1}}{n!}x^n &=\sqrt{x}\small\left(1+\frac3{8x}+\frac1{128x^2}+\frac9{1024x^3}+\frac{491}{32768x^4}+O\left(\frac1{x^5}\right)\right)\tag{11} \end{align} $$

Answer 2

An asymptotic expansion for your series $\;\displaystyle S (x) := e^{-x} \sum_{k=0}^{\infty} \sqrt{k+1}\frac{x^k}{k!} \;$ seems to be, as $\,x\to +\infty$ : $$S(x)\sim\sqrt{x}\left(1+\frac3{8\;x}+\frac 1{128\;x^2}+\frac 9{1024\;x^3}+O\left(\frac 1{x^4}\right)\right)$$ I have no proof for that sorry... (the ideas used are similar to those from this thread).