TITLE: The series $\sum_{n=1}^{\infty} x^{\sum_{k=1}^{n} 1/k}$ converge?
QUESTION [6 upvotes]: The series $\sum_{n=1}^{\infty} x^{\sum_{k=1}^{n} 1/k}$ converge?
For $x>0$
I was looking at some things and I noticed that when $x>1$
$$\lim x^{\sum_{k=1}^{n}1/k} = \infty$$ And therefore, the series in is divergent. The problem is that I do not know very well if the series converges or diverges when $0<x<1$.
Appreciate your help!

REPLY [8 votes]: Hint. Since
$$
\sum_{k=1}^n\frac{1}{k}\approx \log n,
$$
and $x^{\log n}=n^{\log x}$, the sum
$$
\sum_{n=1}^\infty x^{\sum_{k=1}^n\frac{1}{k}}
$$
behaves like the sum
$$
\sum_{n=1}^\infty n^{\log x},
$$
which converges when $\log x<-1$ or $0<x<1/e$.