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Is there a closed form for the series $\sum_{n=0}^{\infty} \frac{(-1)^n}{n!}\mathrm{e}^{-n \theta}$ ?

The series is convergent for all real values of $\theta$ as $\lim_{n \to \infty} \frac{\mathrm{e}^n}{n!}=0$.

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    $\begingroup$ It's just $\exp (-e^{-\theta})$. $\endgroup$
    – lulu
    Commented Jun 27, 2018 at 10:45

1 Answer 1

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$\sum_{n=0}^{\infty} \frac{(-1)^n}{n!}\mathrm{e}^{-n \theta}=\sum_{n=0}^{\infty} \frac{(-e^{-\theta})^n}{n!}= e^{-e^{- \theta}}$.

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