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!


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 $$ \sqrt{x}\le e^{-x}\sum_{k=0}^\infty\sqrt{k+1}\frac{x^k}{k!}\le\sqrt{x+1}\tag{6} $$

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: $$ \begin{align} e^{-x}\sum_{n=1}^\infty\frac{\sqrt{n+1}}{n!}x^n &=\sqrt{x}\small\left(1+\frac3{8x}+\frac1{128x^2}+\frac9{1024x^3}+\frac{491}{32768x^4}+O\left(\frac1{x^5}\right)\right)\tag{11} \end{align} $$

| cite | improve this answer | |
  • $\begingroup$ That's an excellent result. I had suspected the solution (on a hunch), but I couldn't think of how to show it. Also, I've never come across the series form of Jensen's inequality before, so thanks for that too. $\endgroup$ – Donagh Sep 5 '13 at 9:17
  • 1
    $\begingroup$ @Donagh: The requirements of Jensen's Inequality are that the measure space have unit total measure and that the measure be non-negative. $e^{-x}\sum\limits_{k=0}^\infty\frac{x^k}{k!}=1$ and $e^{-x}\frac{x^k}{k!}\ge0$ for $x\ge0$. $\endgroup$ – robjohn Sep 5 '13 at 15:37
  • $\begingroup$ That's great, thanks. I'd come across an integral version before (Gradshteyn and Ryzhik, 12.411), but hadn't realised it was generalisable in this way. I come from an engineering background, so my knowledge of the connections between these things is sometimes a bit limited. $\endgroup$ – Donagh Sep 5 '13 at 17:18

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).

| cite | improve this answer | |
  • $\begingroup$ I was finally able to get your asymptotic expansion (+1) $\endgroup$ – robjohn Sep 6 '13 at 17:06
  • $\begingroup$ @robjohn: Using the central binomial coefficient for that is really neat (+1 of course !). I considered this problem from the point of view of a $\frac 12$- fractional derivative of exponential and this makes your derivation even more interesting! Cheers, $\endgroup$ – Raymond Manzoni Sep 6 '13 at 20:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.