# Power series convergence to an exponential function

Can someone help me to show the proof of the following:

$$\sum^{\infty}_{n=1} \big(\frac{a^n}{n!}\big)^{1/2} \le 2 e^{a}.$$

where $a \ge 0$.

Any help appreciated. Thanks.

• Just square the LHS and exploit the log-convexity of the $\Gamma$ function. Nov 6, 2017 at 9:41
• @JackD'Aurizio Thank you for the hint. Could you please elaborate on log-convexity of $\Gamma$ function. This part is new to me. Detailed steps would help a lot.
– ems
Nov 6, 2017 at 16:27
• You should say $a\ge 0.$
– zhw.
Nov 6, 2017 at 23:00
• @zhw. Thanks. Clarified. Any help with the proof, please?
– ems
Nov 6, 2017 at 23:24

Following the approach suggested in the hint (and assuming $a>0$),

$$\left[\sum_{n\geq 1}\sqrt{\frac{a^n}{n!}}\right]^2 =\sum_{n,m\geq 1}\frac{a^{\frac{m+n}{2}}}{\sqrt{m!n!}}=\sum_{h\geq 2}a^{h/2}\sum_{k=1}^{h-1}\frac{1}{\sqrt{k!(h-k)!}}$$ and by the Cauchy-Schwarz inequality and the (midpoint-)log-convexity of $\Gamma$ we have $$\sum_{k=1}^{h-1}\frac{1}{\sqrt{k!(h-k)!}}\leq\sqrt{(h-1)\frac{2^h-2}{h!}}\leq \frac{2^{h/2}}{(h/2)!}$$ hence

$$\left[\sum_{n\geq 1}\sqrt{\frac{a^n}{n!}}\right]^2\leq \sum_{h\geq 2}\frac{(2a)^{h/2}}{(h/2)!}\leq e^{2a}\left[1+\text{Erf}(\sqrt{2a})\right]\leq 2e^{2a}$$ and by extracting the square root of both sides $$\boxed{ \sum_{n\geq 1}\sqrt{\frac{a^n}{n!}} \leq \color{red}{\sqrt{2}\,e^a}.}$$

Explanations: the first equation is just a rearrangement of an absolutely convergent double series. The Cauchy-Schwarz inequality is used in the form $$\sum a_n \leq \sqrt{\sum 1 \sum a_n^2}$$ and the log-convexity of $\Gamma$ is used to state $\frac{h-1}{h!}\leq\frac{1}{(h/2)!^2}$. At last, the series $\sum_{h\geq 2}\frac{z^{h/2}}{(h/2)!}$ has a closed form only depending on the exponential function and the error function (which is bounded by $1$). By disregarding negative terms and simplifying we reach the highlighted conclusion. Actually a more accurate estimation of $\sum_{k=1}^{h-1}\frac{1}{\sqrt{k!(h-k)!}}$ (for instance through the central limit theorem) allows us to prove the sharper inequality $$\sum_{n\geq 1}\sqrt{\frac{a^n}{n!}} \leq \color{red}{e^a}.$$

• Thanks for your valuable answer @JackD'Aurizio. Tbh, I find each step of yours difficult to understand due to 'my limited math skills'. Do you mind inserting more details into how you go to each next step of equality/inequality? I believe this would also help reach a wider math audience.
– ems
Nov 8, 2017 at 15:57
• @ems: I moved the requested explanations into the answer above. Nov 8, 2017 at 16:09
• Thanks @JackD'Aurizio. There are many invaluable math tricks to learn from your answer and moreover you have a much tighter inequality. Insightful!
– ems
Nov 8, 2017 at 16:30

Since $\sum\limits_{n=2}^\infty\frac1{n(n-1)}=1$, we can use Jensen's Inequality to get a better bound: \begin{align} a^{1/2}+\sum_{n=2}^\infty\left(\frac{a^n}{n!}\right)^{1/2} &=a^{1/2}+\sum_{n=2}^\infty\frac1{n(n-1)}\left(\frac{n(n-1)\,a^n}{(n-2)!}\right)^{1/2}\\ &\le a^{1/2}+\left(\sum_{n=2}^\infty\frac1{n(n-1)}\frac{n(n-1)\,a^n}{(n-2)!}\right)^{1/2}\\ &=a^{1/2}+\left(\sum_{n=2}^\infty\frac{a^n}{(n-2)!}\right)^{1/2}\\[6pt] &=a^{1/2}+ae^{a/2}\\[12pt] &\lt2e^a-\frac74 \end{align}

For the last inequality, because $\frac a2\le e^{a/2}-1$, we have \begin{align} 2e^a-ae^{a/2}-a^{1/2} &\ge2e^a-2\!\left(e^{a/2}-1\right)\!e^{a/2}-a^{1/2}\\[3pt] &=2e^{a/2}-a^{1/2}\\ &\ge\operatorname{W}\!\left(\frac14\right)^{-1/2}\!\!-\,\,\operatorname{W}\!\left(\frac14\right)^{1/2}\\[3pt] &\doteq1.76310333146\\ &\gt\frac74 \end{align} since $\frac{\mathrm{d}}{\mathrm{d}a}\left(2e^{a/2}-a^{1/2}\right)=e^{a/2}-\frac1{2a^{1/2}}=0$ when $a=\operatorname{W}\!\left(\frac14\right)$, where $\operatorname{W}$ is Lambert W.

• That's such an elegant proof. Can you elaborate on how you obtain the first inequality? I believe it has to do with Jensen's inequality, but cannot see it clearly. I look forward to your clarification with eager.
– ems
Nov 6, 2017 at 23:50
• Also, I do not see where and how you have used the first convergence result (i.e., $\sum\limits_{n=2}^\infty\frac1{n(n-1)}=1$) in the proof. Please clarify.
– ems
Nov 6, 2017 at 23:57
• Jensen's inequality says that, for a convex function $f$, if $$\sum_{n=2}^\infty w_n=1$$ then $$\sum_{n=2}^\infty w_nf\left(x_n\right) \ge f\!\left(\sum_{n=2}^\infty w_nx_n\right)$$ In the proof, we use $w_n=\frac1{n(n-1)}$, $f(x)=x^2$, and $x_n=\left(\frac{n(n-1)\,a^n}{(n-2)!}\right)^{1/2}$.
– robjohn
Nov 6, 2017 at 23:59
• Actually the constant $\color{red}{2}$ can be lowered down to $\sqrt{2}$ with elementary methods and down to $1$ with probabilistic arguments. Nov 8, 2017 at 16:18
• @JackD'Aurizio: Indeed, $a^{1/2}+ae^{a/2}\lt e^a$, so the bound shown above shows that, too. I've plotted it, but I will work on a proof.
– robjohn
Nov 8, 2017 at 17:12