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How to determine the value of $\displaystyle f(x) = \sum_{n=1}^\infty\frac{\sqrt n}{n!}x^n$? No context, this is just a curiosity o'mine.

Yes, I am aware there is no reason to believe a random power series will have a closed form in terms of well established functions, but also I have no way to know if that is the case here, so that is why I'm asking. Do you know this power series or any method I could use to determine its value?

In my research I've found out about the polylogarithm, which is defined as $$\mathrm{Li}_s(x) = \sum_{n=1}^\infty\frac{x^n}{n^s} = \frac1{\Gamma(s)}\int_0^\infty\frac{t^{s-1}}{e^t/x-1}dt$$

This called my attention because $$\begin{aligned} f(x) &= \sum_{n=1}^\infty\frac{\sqrt n}{n!}x^n\\ &= x\sum_{n=1}^\infty\frac1{\sqrt n}\frac{x^{n-1}}{(n-1)!}\\ &= x\sum_{n=1}^\infty\frac1{\sqrt n}\mathcal L^{-1}\left\{\frac1{x^n}\right\}\\ &= x\mathcal L^{-1}\left\{\sum_{n=1}^\infty\frac1{\sqrt n}\frac1{x^n}\right\}\\ &= x\mathcal L^{-1}\left\{\mathrm{Li}_{1/2}\left(\frac 1x\right)\right\} \end{aligned}$$

Yeah... This is not the closed form I was expecting. Can we do better?


A comment suggested this post might have the answer to my question. However, despite it asking for the same thing, it asks as well for similar and more general expressions, which motivates less specific responses. In fact, the accepted (and only) answer in the post discusses only the asymptotic behavior of the series I'm interested in.

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    $\begingroup$ Notice how this looks similar to taylor series, here we need to find a function such that $f^{(n)}(0)=\sqrt n$. I haven't found the answer yet though. $\endgroup$ Commented Mar 26 at 17:25
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    $\begingroup$ @ThomasAndrews Read the second paragraph in the OP. $\endgroup$ Commented Mar 26 at 17:28
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    $\begingroup$ The Bell polynomial $B_n(x)$ would work if $n$ could be $\in\Bbb Q$. $\endgroup$ Commented Mar 26 at 17:29
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    $\begingroup$ According to the paper Computation of certain infinite series of the form $\sum f(n)n^k$ for arbitrary real-valued $k$, we have $$\sum_{n=0}^\infty\frac{\sqrt n}{n!}=\frac1{\sqrt\pi}\int_0^\infty\frac{e^{e^{-x}-x}}{\sqrt x}~{\rm d}x$$ $\endgroup$ Commented Mar 26 at 19:01
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    $\begingroup$ $$\sum_{n=1}^\infty\frac{\sqrt n}{n!}x^n=x\sum_{n=0}^\infty \frac{x^n}{n!\sqrt {n+1}}=x\sum_{n=0}^\infty \frac{x^n}{n!}\frac1{\sqrt\pi}\int_0^\infty t^{-1/2}e^{-(n+1)t}dt=\frac x{\sqrt\pi}\int_0^\infty e^{xe^{-t}-t}\frac{dt}{\sqrt t}$$ $\endgroup$
    – Svyatoslav
    Commented Mar 27 at 1:15

2 Answers 2

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In my opinion, getting the integral representation is too little for a post, but in order to put some meat on the bones, we can find a full asymptotics using this formula. $$S(x)=\sum_{n=1}^\infty\frac{\sqrt n}{n!}x^n=x\sum_{n=0}^\infty \frac{x^n}{n!\sqrt {n+1}}=x\sum_{n=0}^\infty \frac{x^n}{n!}\frac1{\sqrt\pi}\int_0^\infty t^{-1/2}e^{-(n+1)t}dt$$ Performing summation first, $$S(x)=\frac x{\sqrt\pi}\int_0^\infty e^{xe^{-t}-t}\frac{dt}{\sqrt t}$$ The asymptotics at $x\to0$ is evident. Decomposing $e^{xe^{-t}}$ near $x=0$ $$S(x)\sim\frac x{\sqrt\pi}\int_0^\infty e^{-t}\big(1+xe^{-t}+\frac{x^2}{2!}e^{-2t}+...\big)\frac{dt}{\sqrt t}$$ what, of course, after integration simply coincides with the initial sum.

At $x\to\infty$ $$S(x)\overset{s=e^{-t}}{=}\frac x{\sqrt\pi}\int_0^1\frac{e^{xt}}{\sqrt{\ln\frac1t}}dt\overset{t=1-s}{=}\frac{xe^x}{\sqrt\pi}\int_0^1\frac{e^{-xs}}{\sqrt{\ln\frac1{1-s}}}ds\overset{xs=t}{=}\frac{e^x}{\sqrt\pi}\int_0^x\frac{e^{-t}}{\sqrt{\ln\frac1{1-\frac tx}}}dt$$ The integrand is declining when $t$ is growing, and becomes exponentially small, for example, at $t\sim\sqrt x$. It means that we are allowed to decompose the denominator near $t=0$ and integrate term by term. Expanding integration to $\infty$ and dropping exponentially small terms $$S(x)\sim\frac{e^x}{\sqrt\pi}\int_0^x\frac{e^{-t}}{\sqrt{\frac tx+\frac{t^2}{2x^2}+...}}dt\sim\frac{\sqrt x\,e^x}{\sqrt\pi}\int_0^\infty e^{-t}\Big(1-\frac t{4x}-\frac7{96}\frac1{x^2}+...\Big)\frac{dt}{\sqrt t}$$ $$S(x)=\sqrt x\,e^x\left(1-\frac1{8x}-\frac7{128x^2}+O\Big(\frac1{x^3}\Big)\right)$$

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    $\begingroup$ Is there an explicit infinite sum for the final formula? $\endgroup$ Commented Mar 27 at 12:21
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    $\begingroup$ Nice derivation of the asymptotic for $x\to \infty$. For the more general sum $s(a,x) = \sum_{n=1}^\infty\frac{n^a}{n!}x^n$ ($0 \lt a \le1$ ) we find the leading term $s(a,x) \to x^a e^{x}$. Could this simple result perhaps be found more easily? $\endgroup$ Commented Mar 28 at 8:38
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    $\begingroup$ @Svyatoslav For a non-negative integer $r$ we have $\sum _{n=0}^{\infty } \frac{n^r x^n}{n!}=e^x B_r(x)$. Where $B_r(x)$ is the Benoulli polynom given by $B_r(x) = \sum _{k=0}^r x^k \mathcal{S}_r^{(k)}$ where $\mathcal{S}_r^{(k)}$ are the Stirling numbers of the second kind. $\endgroup$ Commented Mar 28 at 18:35
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    $\begingroup$ @Svyatoslav For any $a \ge 0$ I find (see also my solution) $s(a,x) := \sum_{n=0}^{\infty} \frac{n^a}{n!}x^n = \frac{x}{\Gamma(1-\{a\})} \frac{\partial ^{\lfloor a\rfloor}}{\partial w^{\lfloor a\rfloor}}\left(\int_0^1 \frac{\exp \left(-u x\; e^w +e^w x+w\right)}{\log ^{\{a\}}\left(\frac{1}{1-u}\right)} \, du\right)|_ {w\to 0}$ $\endgroup$ Commented Mar 29 at 10:46
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    $\begingroup$ Alternative approach: math.stackexchange.com/q/378528 $\endgroup$
    – Gary
    Commented Mar 29 at 11:18
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This is not a complete solution but it gives only the result: a complete formula for a sum generalizing that of the OP.

The development was initiated by the results of @Svyatoslav and completed to this point in a discssion with him.

$$s(a,x) := \sum_{n=0}^{\infty} \frac{n^a}{n!}x^n \\ = \frac{x}{\Gamma(1-\{a\})} \frac{\partial ^{\lfloor a\rfloor}}{\partial w^{\lfloor a\rfloor}}\left(\int_0^1 \frac{\exp \left(-u x\; e^w +e^w x+w\right)}{\log ^{\{a\}}\left(\frac{1}{1-u}\right)} \, du\right)|_ {w\to 0}$$

Here $\lfloor a\rfloor$ is the integer part of $a$ and $\{a\} = a-\lfloor a\rfloor$ is the fractional part of $a$.

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