Limit of $\sum_1^\infty n^\alpha e^{- n^\alpha}$

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 $$\int_0^1 \sum_1^\infty n^\alpha e^{-t n^\alpha} \,\mathrm{d}t$$

• 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. Commented May 8, 2019 at 2:19
• In closed form? Almost certainly not in general, although the case $\alpha=2$ can be expressed in terms of Jacobi theta functions. Commented May 8, 2019 at 2:21
• @RobertIsrael in fact, an upper bound might be enough (see my edit concerning what I truly want to achieve) Commented May 8, 2019 at 2:33
• 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. Commented May 8, 2019 at 3:07
• Integrate first, then sum. But it isn't finite. Commented May 8, 2019 at 12:17