how to find the series $x + x^{1 + \frac{1}{2}} + x^{1 + \frac{1}{2}+ \frac{1}{3}} +...$ is convergent. The series $x + x^{1 + \frac{1}{2}} + x^{1 + \frac{1}{2}+ \frac{1}{3}} +...$
is convergent if 
(A) $x>e$ 
(B) $x<e $
(C) $x<1/e$ 
(D) $x>1/e$
I think the answer is C, but I could not determine the condition... how to solve it. Plz help.
 A: We have
\begin{align*}
  x^{\sum_{i=1}^n \frac 1i} &\le x^{\log n + 1}\\
   &= x \cdot \exp(\log n\cdot \log x)\\
   &= x \cdot n^{\log x}
\end{align*}
and
\begin{align*}
  x^{\sum_{i=1}^n \frac 1i} &\ge x^{\log n}\\
   &= \exp(\log n\cdot \log x)\\
   &= n^{\log x}
\end{align*}
and hence
\[ \sum_{n=1}^\infty x^{\sum_{i=1}^n \frac 1i} < \infty \iff \sum_{n=1}^\infty n^{\log x} < \infty \]
which is true exactly for $\log x < -1\iff x < \frac 1e$.
A: Nth term of series is given by $u_n=x^{\sum_{i=1}^n \frac{1}{i}}$
Logarithmic test says

Suppose $\sum_{ n \geq 1} a_n $ is a series of positive terms. Suppose
that
$$ \lim_{n \to \infty} n \ln \dfrac{ a_n }{a_{n+1} } = g $$
and if $g>1$ series is convergent and divergent if $g<1$.

now firstly take $x\ge0$, at $x=0$ series converges, and at $x=1$ series diverges(option B discarded).
Now $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4} \cdots$ is divergent series, so $\lim( 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4} \cdots)$ tends to infinity so if $x>1$, $lim(x^{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4} \cdots}) \ne 0$, so not convergent for $x>1$(option A discarded).
applying log test for $0<x<1$, $$\lim_{n \to \infty} n \log \dfrac{ a_n }{a_{n+1} }=n \log(x^{\frac{-1}{n}})=-\log x$$
so if $-\log x > 1$ or $x<\frac{1}{e}$ , series converges and if $x>\frac{1}{e}$ series diverges.
For negative values of $x$ , series is absolutely convergent for $x<\frac{1}{e}$ so it is convergent in $x<\frac{1}{e}$ region.
