<s>[INST] How to determine whether this series convergent or divergent? Does
$$
\dfrac{7}{19}+\dfrac{7}{19}\sqrt{\dfrac{7}{19}}+\dfrac{7}{19}\sqrt{\dfrac{7}{19}}\sqrt[3]{\dfrac{7}{19}}+\cdots+\dfrac{7}{19}\sqrt{\dfrac{7}{19}}\sqrt[3]{\dfrac{7}{19}}\cdots\sqrt[n]{\dfrac{7}{19}}+\cdots
$$ 
converge or diverge?
The following is my idea:
use
$1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}>\ln{n}$
$$
  \sum_{n=1}^{\infty}\left(\dfrac{7}{19}\right)^{(1+1/2+\cdots+1/n)}
< \sum_{n=1}^{\infty}\left(\dfrac{7}{19}\right)^{\ln{n}}
= \sum_{n=1}^{\infty}n^{-\ln(19/7)}
$$
But $p=\ln{\dfrac{19}{7}}<1$, becasue $\dfrac{19}{7}\approx 2.71428<e=2.71828$
I guess the series is divergent because I use 
$$1+1/2+\cdots+1/n\approx \ln{n}, n\to\infty$$
to find $$\sum_{n=1}^{\infty} (1/x)^{1+1/2+1/3+\cdots+1/n}$$ is convergent only if $x>e$.
So, my question is: how do I determine whether
$$
\dfrac{7}{19}+\dfrac{7}{19}\sqrt{\dfrac{7}{19}}+\dfrac{7}{19}\sqrt{\dfrac{7}{19}}\sqrt[3]{\dfrac{7}{19}}+\cdots+\dfrac{7}{19}\sqrt{\dfrac{7}{19}}\sqrt[3]{\dfrac{7}{19}}\cdots\sqrt[n]{\dfrac{7}{19}}+\cdots
$$ 
converges or diverges? Thank you 
 [/INST] Since
$$
H_n = \sum_{k=1}^{n} \frac{1}{k} = \log n + \gamma + o(1),
$$
we have, for $x > 0$,
$$
x^{H_n} \sim x^{\log n + \gamma} = x^\gamma n^{\log x}
$$
As $n \to \infty$.  We may then apply the limit comparison test to conclude that
$$
\sum_{n=1}^{\infty} \frac{1}{x^{H_n}}
$$
converges when $\log x > 1 \Leftrightarrow x > e$ and diverges when $\log x \leq 1 \Leftrightarrow x \leq e$.
Since $19/7 < e$ the series in your question diverges.
</s>