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I know that $\sum_{k=0}^\infty \dfrac{k^k}{k!}\dfrac{e^{-k-k\tfrac{s}{n}}}{\sqrt{n}}$ can be expanded in powers of $\frac{1}{\sqrt{n}}$ to yield:

$\frac1{\sqrt{2s}}+\frac{1}{3\sqrt{n}}+\frac{\sqrt{2s}}{12n}+\dotso$

My question is twofold: (i) is there an actual proof of this (beyond numerical), (ii) how does this expansion continue (general formula, or at least the next few terms)?

Honza
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    Here is how to use MathJax. – Toby Mak Jun 01 '19 at 00:02
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    I edited your question. Please check if this is how you meant it and let me know, if there is a mistake in it. – Cornman Jun 01 '19 at 00:04
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    Why doesn't the expression in the title have $n$ in it? If you are interested in the expansion in terms of $n$, then you don't need the sum over $k$ at all, you only need to leave the general term containing $n$ – Yuriy S Jun 01 '19 at 00:14

2 Answers2

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We will assume $s > 0$ or the series diverges.

Recall for $|z| \le \frac1e$, the $0$-branch of the Lambert W function has following expansion:

$$W_0(z) = \sum_{k=1}^\infty \frac{(-k)^{k-1}}{k!} z^k$$

Pick any $\lambda_0 > 1$ and substitute $z$ by $-e^{-\lambda}$ for $\lambda > \lambda_0$. Above expansion converges uniformly for such $\lambda$. So does the series of its derivatives for each individual term. In this situation, it is legal to exchange the order of summation and taking derivatives. This implies

$$f(\lambda) \stackrel{def}{=}\sum_{k=0}^\infty \frac{k^k}{k!} e^{-\lambda k} = 1 + \sum_{k=1}^\infty \frac{k^k}{k!} e^{-\lambda k} = 1 + \frac{d}{d\lambda} W_0(-e^{-\lambda}) $$ Let $u = W_0(-e^{-\lambda})$, we have $u e^u = -e^{-\lambda}$. Differentiate against $\lambda$, we get $$(u+1)e^u \frac{du}{d\lambda} = e^{-\lambda} = -u e^u \implies \frac{du}{d\lambda} = - \frac{u}{1+u}$$ This leads to

$$f(\lambda) = 1 + \frac{du}{d\lambda} = \frac{1}{1+u} = \frac{1}{1 + W_0(-e^{-\lambda})}$$

It is easy to see the series at hand evaluates to $\displaystyle\;\frac{1}{\sqrt{n}}f\left(1 + \frac{s}{n}\right)$

For simplicity, let's assume $n = 1$. Let $\lambda = 1 + s$ and $w = 1 + u$. We have $$f(1 + s) = \frac{1}{w}\quad\text{ and }\quad ue^u = -e^{-\lambda} \iff (1-w) e^w = e^{-s}$$ This leads to $$s = -(\log(1-w) + w) = \frac{w^2}{2} + \frac{w^3}{3} + \frac{w^4}{4} + \cdots $$ For small $s$, the leading approximation of $w$ is clearly $\sqrt{2s}$ and hence $f(1+s) \sim \frac{1}{\sqrt{2s}}$.

To process further, write $s$ as $\frac{t^2}{2}$ and expanse $w$ as a power series in $t = \sqrt{2s}$. By matching coefficients of different powers of $t$ using a CAS, we get

$$\begin{align}w &= t-\frac{t^2}{3}+\frac{t^3}{36}+\frac{t^4}{270}+\frac{t^5}{4320}-\frac{t^6}{17010} + \cdots\\ \implies \frac1w &= \frac{1}{t}+\frac{1}{3}+\frac{t}{12}+\frac{2t^2}{135}+\frac{t^3}{864}-\frac{t^4}{2835}-\frac{139 t^5}{777600} + \cdots \end{align} $$ Putting $n$ back, we get

$$\frac{1}{\sqrt{n}}f\left(1 + \frac{s}{n}\right) = \frac{1}{\sqrt{2s}} + \frac{1}{3n^{\frac12}} + \frac{\sqrt{2s}}{12n} +\frac{4\,s}{135\,{n}^{\frac{3}{2}}} +\frac{\sqrt{2s}^3}{864 n^2} -\frac{4\,{s}^{2}}{2835\,{n}^{\frac{5}{2}}} + \cdots $$ It looks like there is a typo in the third term of the expansion you get.

achille hui
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Since $k! \le ek^{k+\frac{1}{2}} e^{-k}$ for integers $k \ge 1$, we have \begin{align*} \sum_{k=0}^{\infty} \frac{k^k}{k!}e^{-k} \ge \sum_{k=1}^{\infty} \frac{1}{e\sqrt{k}}=\infty \end{align*} where I assumed the notion $0^0 = 0$. The divergence is the same if we assumed $0^0 = 1$.

Tom Chen
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