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I want to find the residue of $$ \sum_{n=1}^{\infty} \frac{z^n}{n!(1-z^n)} $$ at $z=1$. I've tried $$ \sum_{n=1}^{\infty} \frac{z^n}{n!(1-z^n)} = \sum_{n=1}^{\infty} \left( \frac{1}{n!(1-z^n)} - \frac{1}{n!} \right) = 1-e +\sum_{n=1}^{\infty} \frac{1}{n!(1-z^n)} $$ but I don't really see how to get the summand into a multiple of $(z-1)^n$ after this.

Rowan
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    $1-z^n = (1-z)(\sum z^k)$. – Robert Shore Jun 22 '19 at 01:13
  • This series has convergence issues. Define $w_k=\exp(2\pi i/k)$. Then the $k$-th term of your series will have $1-w_k^k=0$ in the denominator. Since the $w_k$ come arbitrarily close to $z=1$, it isn't possible to just cut out a small disk around $z=1$ to avoid this. This isn't to say that the singularities aren't removable, but convergence certainly has to be investigated carefully. If they turn out not to be removable, the function can't be holomorphic at $1$ even after multiplication by $(1-z)$ – J_P Jun 22 '19 at 01:34
  • @RobertShore $z^n-1$ has $n$ roots, not only the root at $z=1$. – Mark Viola Jun 22 '19 at 02:48

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