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 

It can be used Raabe's test.
We have series
$$\sum_{n=1}^{\infty} a_n, \qquad \mathrm{where } \quad
a_n = \left(\dfrac{7}{19}\right)^{1+\frac{1}{2}+\cdots+\frac{1}{n}}.
$$
We wil construct value $R_n= n \left( \dfrac{a_{n+1}}{a_n}-1\right)$. Denote 
$R=\lim\limits_{n\to\infty} R_n$.
If $R<-1$, then series converges.
If $R>-1$, then series diverges.
$R_n = n\left(\left(\dfrac{7}{19}\right)^{\frac{1}{n+1}}-1\right)= n\left( \exp(\frac{1}{n+1}\ln\frac{7}{19})-1\right)$.
Using Taylor series for $\exp$, we get
$$R_n = n \sum_{j=1}^{\infty}  \dfrac{ \left(\frac{1}{n+1} \ln\frac{7}{19}\right)^j }{j!}.$$
So, $R=\lim\limits_{n\to\infty} R_n = \ln\frac{7}{19} >\ln \frac{1}{e}=-1$. (because 
$\frac{19}{7}<e$). 
Actually, $R \approx -0,99852883011112715490367468844.\;$
So, series diverges.
