Find if $\sum\limits_{n=1}^{\infty} a^{1+\frac1{2}+\frac1{3}+\dots+\frac1{n}}$, $a > 0$ converges or not.

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 $a\ge 1/e$.

hint

You can use the fact that the sequence $$\sum_{k=1}^n\frac 1k-\ln(n)$$

converges to Euler's constant $\gamma$.