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I came across this nice equality that I don't know how to solve:

$$\sum_{j=1}^{\infty} \frac{e^{-j}j^{j-1}}{j!}=1.$$

I'm guessing the first step is to write out the Taylor series, which gives a double sum of:

$$\sum_{j=1}^{\infty} \frac{\left(\sum_{n=1}^{\infty}\frac{(-j)^n}{n!}\right)j^{j-1}}{j!}$$

and maybe this can be seen to be the product of two other functions in Taylor form, which I can't see. Also, I know this can be solved by elementary methods, so a "simple" solution would be nice. Help is appreciated!

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  • $\begingroup$ This is an equality, not an inequality. $\endgroup$
    – robjohn
    Oct 19, 2016 at 10:08
  • $\begingroup$ $x$ should read $j$? $\endgroup$ Oct 19, 2016 at 13:04

4 Answers 4

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Hint. For $|x|< 1/e$, the power series $$T(x)=\sum_{j=1}^{\infty}\frac{j^{j-1}}{j!}\,x^j,$$ which is the exponential generating function for rooted labeled trees, satisfies the identity $$T(x)=xe^{T(x)}.$$ Now you need to show that $T(1/e)=1$.

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By Lagrange inversion theorem, the principal branch of the Lambert $W$ function, defined through $W(x)e^{W(x)}=x$, has the following expansion near the origin: $$ W(x) = \sum_{n\geq 1}\frac{n^{n-1}(-1)^{n-1}}{n!}x^n\tag{1} $$ In particular, $$ \sum_{n\geq 1}\frac{n^{n-1}}{n! e^{n}} = -W\left(-\frac{1}{e}\right) = \color{red}{1}\tag{2} $$ since the function $x\cdot e^x$ equals $-\frac{1}{e}$ at $x=-1$, from which $W\left(-\frac{1}{e}\right)=-1$.

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    $\begingroup$ This is very interesting ! In spite of my love for Lambert function, I missed it. Thanks for posting this nice solution. $\endgroup$ Oct 19, 2016 at 18:01
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    $\begingroup$ This is the answer I was trying to find (+1) I had proven $(1)$ when this question was asked in chat. $\endgroup$
    – robjohn
    Oct 20, 2016 at 17:10
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You can get an approximate value of $$\sum_{n=1}^\infty\frac{e^{-n}\, n^{(n-1)}}{n!}$$ using Stirling approximation of the factorial.

Since $$n! \sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n \left(1 +\frac{1}{12n}+\frac{1}{288n^2} - \frac{139}{51840n^3} -\frac{571}{2488320n^4}+ \cdots \right)$$ that, for the time being we can write, $$a_n=\frac{e^{-n}\, n^{(n-1)}}{n!}\sim\frac{1}{\sqrt{2 \pi } n^{3/2} P_n}$$ where $P_n$ is the term in brackets.

Now, using long division (or, better, Taylor series) $$\frac{1}{P_n}=1-\frac{1}{12 n}+\frac{1}{288 n^2}+\frac{139}{51840 n^3}+\frac{571}{311040 n^4}+O\left(\frac{1}{n^5}\right)$$ which makes $$a_n\sim \frac{1}{\sqrt{2\pi}}\left(\frac 1 {n^{3/2}} -\frac 1 {12} \frac 1 {n^{5/2}}+\frac1 {288} \frac 1 {n^{7/2}}+\frac{139} {51840} \frac 1 {n^{9/2}}+\frac{571} {311040} \frac 1 {n^{11/2}} \right)+\cdots$$ Now summation $$\sum_{n=1}^\infty a_n\sim \frac{1}{\sqrt{2\pi}}\left(\zeta \left(\frac{3}{2}\right)-\frac{\zeta \left(\frac{5}{2}\right)}{12}+\frac{\zeta \left(\frac{7}{2}\right)}{288}+\frac{139 \zeta \left(\frac{9}{2}\right)}{51840}+\frac{571 \zeta \left(\frac{11}{2}\right)}{311040} \right)\approx 1.00103$$ Adding more terms will make the result closer and closer to $1$.

Edit

If, instead of Stirling, we use Ramanujan's famous factorial approximation, $$n!\approx\sqrt{\pi}\left(\frac{n}{e}\right)^n\root\LARGE{6}\of{8n^3+4n^2+n+\frac{1}{30}}$$ and expand the radical as a Taylor series, we should arrive to $$a_n\sim \frac{1}{\sqrt{2\pi}}\left(\zeta \left(\frac{3}{2}\right)-\frac{\zeta \left(\frac{5}{2}\right)}{12}+\frac{\zeta \left(\frac{7}{2}\right)}{288}+\frac{139 \zeta \left(\frac{9}{2}\right)}{51840}-\frac{2947 \zeta \left(\frac{11}{2}\right)}{2488320} \right)\approx 0.99990$$

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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{j\ =\ 1}^{\infty}{\expo{-j}j^{\, j - 1} \over j!} & = \sum_{j\ =\ 1}^{\infty}{\expo{-j} \over j!}\,\,\, \overbrace{\pars{j - 1}! \oint_{\verts{z}\ =\ 1}{\expo{\, jz} \over z^{\, j}}\,{\dd z \over 2\pi\ic}} ^{\ds{=\ j^{\, j - 1}}}\,\,\, =\,\,\, \oint_{\verts{z}\ =\ 1}\,\,\,\sum_{j\ =\ 1}^{\infty} {\pars{\expo{z - 1}/z}^{\, j} \over j}\,{\dd z \over 2\pi\ic} \\[5mm] & = -\oint_{\verts{z}\ =\ 1}\,\,\, \ln\pars{1 - {\expo{z - 1} \over z}}\,{\dd z \over 2\pi\ic} = -1 - \oint_{\verts{z}\ =\ 1}\,\,\, \ln\pars{z - \expo{z - 1}}\,{\dd z \over 2\pi\ic} \end{align} The equation $\ds{z - \expo{z - 1} = 0}$ has infinite solutions$^{\large\S}$. One of them is obviously $\ds{z = 1}$ and the remaining ones have $\ds{\verts{z} > 1}$. $$\bbox[#ffd,10px,border:0.5px groove navy] {\begin{array}{l} {\large\S}:~\mbox{They are evaluated by}\ \texttt{Mathematica}\ \mbox{as}\ \texttt{ProductLog[k,-1/e]}\ \mbox{with}\,\,\, k \in \mathbb{Z}\ \end{array}} $$ It reduces the complex integration to: \begin{align} \sum_{j\ =\ 1}^{\infty}{\expo{-j}j^{\, j - 1} \over j!} & = -1\ -\,\,\, \overbrace{\oint_{\verts{z}\ =\ 1}\,\,\, \ln\pars{z - 1}\,{\dd z \over 2\pi\ic}}^{\ds{=\ -2}} = \bbx{\ds{1}} \end{align}

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