TITLE: Find if $\sum\limits_{n=1}^{\infty} a^{1+\frac1{2}+\frac1{3}+\dots+\frac1{n}}$, $a > 0$ converges or not.
QUESTION [9 upvotes]: Find if $\sum\limits_{n=1}^{\infty} a^{1+\frac1{2}+\frac1{3}+\dots+\frac1{n}}$, $a >0$ converges or not.
I used d'Alembert's criterion and I found $\lim_{n\to \infty} \frac{x_n}{x_{n+1}}=1$
Moving to Raabe–Duhamel's test, I found $$\lim_{n\to \infty} n(\frac{x_n}{x_{n+1}} -1) =  \lim_{n\to \infty} n(\frac{1}{\sqrt[n+1]{a}} - 1) $$
What can I do from there? Or is there a better way to find if $\sum\limits_{n=1}^{\infty} a^{1+\frac1{2}+\frac1{3}+\dots+\frac1{n}}$, $a >0$ converges or not?

REPLY [17 votes]: First owing to the concavity of the logarithm function, we have
$$\int_1^n \frac1x\,dx\le \sum_{k=1}^n\frac1k \le 1+\int_1^n \frac1x\,dx$$
Therefore, for $1>a>0$ we have
$$a^{\log(n)+1}\le a^{\sum_{k=1}^n \frac1k}\le a^{\log(n)}$$
while for $a\ge 1$ we have
$$a^{\log(n)}\le a^{\sum_{k=1}^n \frac1k}\le a^{\log(n)+1}$$
Thus, using $a^{\log(n)}=n^{\log(a)}$ we find from by "the p-test" that the series converges whenever $0<a<1/e$ and diverges for $a\ge 1/e$.

REPLY [3 votes]: hint
You can use the fact that the sequence
$$\sum_{k=1}^n\frac 1k-\ln(n)$$
converges to Euler's constant $\gamma$.