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So consider the sum $$S=\sum_{m=1}^n(-1)^m\frac{m^{n-m-1}}{(n-m)!},$$ where $n$ is some fixed, positive integer. For specific values of $n$, this gives $-1$, $-\frac{1}{2}$, $\frac{1}{6}$, $\frac{1}{12}$, $-\frac{3}{40}$, $-\frac{1}{120}$, $\frac{31}{1008}$, $-\frac{29}{5040}$, $-\frac{7}{640}$, $\frac{2087}{362880}$,... for $n=1, 2, 3, 4, 5, 6, 7, 8, 9, 10,$ ... Is there a way to evaluate this sum for general $n$?

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  • $\begingroup$ curious, is there any additional context for this? $\endgroup$
    – C Squared
    Apr 18, 2021 at 15:29
  • $\begingroup$ Not especially. It was part of a combinatorial factor that popped out while I was doing some quantum Fermi gas calculations, and I wondered if it had a clean expression. $\endgroup$
    – arow257
    Apr 18, 2021 at 16:02
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    $\begingroup$ It appears $n!S(n)$ is an integer for all $n$. Maybe that integer sequence is in OEIS? $\endgroup$ Apr 18, 2021 at 17:08
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    $\begingroup$ Thanks! I checked OEIS and it seems that the $S(n)$ are the coefficients of $x^n$ in the Taylor expansion of $-\log(1 + x e^x)$. $\endgroup$
    – arow257
    Apr 18, 2021 at 17:33

1 Answer 1

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The generating function for

$$S_n = \sum_{m=1}^n (-1)^m \frac{m^{n-m-1}}{(n-m)!}$$

may be proved as follows:

$$\sum_{m=0}^{n-1} (-1)^{n-m} \frac{(n-m)^{m-1}}{m!} = \frac{(-1)^n}{n} + \sum_{m=1}^{n-1} (-1)^{n-m} \frac{(n-m)^{m-1}}{m!} \\ = \frac{(-1)^n}{n} + \sum_{m=1}^{n-1} (-1)^{n-m} \frac{1}{m} [z^{m-1}] \exp((n-m)z) \\ = \frac{(-1)^n}{n} + \sum_{m=1}^{n-1} (-1)^{n-m} \frac{1}{n-m} [z^m] \exp((n-m)z) \\ = \frac{(-1)^n}{n} + \sum_{m=1}^{n-1} (-1)^{m} \frac{1}{m} [z^{n-m}] \exp(mz) \\ = \sum_{m=1}^{n} (-1)^{m} \frac{1}{m} [z^{n-m}] \exp(mz) \\ = [z^n] \sum_{m=1}^{n} (-1)^{m} \frac{1}{m} z^m \exp(mz).$$

Now here the coefficient extractor enforces the range and we get

$$[z^n] \sum_{m\ge 1} (-1)^{m} \frac{1}{m} z^m \exp(mz) = [z^n] \log\frac{1}{1+z\exp(z)}.$$

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