I want to study the convergence of the series $\sum n^{-\frac{x^n}{n}}$, but none of the criteria I know seem to work. What could I do?

  • 4
    $\begingroup$ How about writing $n^{-x^n/n}$ as $e^{-\frac{x^n}{n}\ln(n)}$? $\endgroup$
    – Enrico M.
    Nov 13, 2022 at 13:44
  • $\begingroup$ If $|x|\le1$ the series cannot converge $\endgroup$
    – FShrike
    Nov 13, 2022 at 13:51

1 Answer 1


The series converges when $x > 1$, and diverges when $x \leq 1$ First note that it trivially doesn't converge for negative $x$ since for those values, the series has terms greater than 1 when $n$ is odd. $x=0$ obviously diverges.

So let us now assume that $x \geq 1$, then for large $n$, $n^{\frac{-x^n}{n}} < \frac{1}{n^2}$, since $x^n$ grows faster than $n^k$ for any $k$. And that series converges.

For $0< x \leq 1$, the terms are going to 1, and so the series diverges.


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