Let $\alpha > 1$. The series $\sum_1^\infty n^\alpha e^{- n^\alpha}$ is convergent (using the ratio test for example).

Is there a way to compute its sum ?

Edit: or maybe a upper bound of its sum ? My ultimate goal is to show the finiteness of $\int_0^1 \sum_1^\infty n^\alpha e^{-t n^\alpha} \,\mathrm{d}t$

  • $\begingroup$ Don't know about the sum, but the integral $\int_0^{\infty} n^{\alpha}e^{-n^{\alpha}}dn= \frac{1}{\alpha}\Gamma\left(\frac{1}{\alpha}+1\right),$ so it's "close" to that. $\endgroup$
    – Dzoooks
    Commented May 8, 2019 at 2:19
  • $\begingroup$ In closed form? Almost certainly not in general, although the case $\alpha=2$ can be expressed in terms of Jacobi theta functions. $\endgroup$ Commented May 8, 2019 at 2:21
  • $\begingroup$ @RobertIsrael in fact, an upper bound might be enough (see my edit concerning what I truly want to achieve) $\endgroup$ Commented May 8, 2019 at 2:33
  • 1
    $\begingroup$ Let $q$ be any integer exceeding $\alpha$. Then your sum is bounded by $\sum n^qe^{-n}$, and that sum can be computed in closed form. $\endgroup$ Commented May 8, 2019 at 3:07
  • $\begingroup$ Integrate first, then sum. But it isn't finite. $\endgroup$ Commented May 8, 2019 at 12:17


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