Is the series convergent $x + x^{1+\frac{1}{2}}+x^{1+\frac{1}{2}+ \frac{1}{3}}+\cdots$ $$x + x^{1+\frac{1}{2}}+x^{1+\frac{1}{2}+ \frac{1}{3}}+\cdots$$
I know that $u_n = \lim_{n \to \infty}   x^ {\sum_{i=1}^n(1/i)}$
I used root test and ratio test but it isn't working.
 A: Partial answer:
The $n$th term is
$$x^{1+{1\over2}+\cdots+{1\over n}}\approx x^{\log n}=e^{\log x\log n}=n^{\log x}$$
so you would expect the series to converge when $\log x\lt-1$ and diverge when $\log x\gt-1$.  It's a little unclear what happens when $\log x=-1$ (since $n^{\log x}$ is only an approximation).
A: Write $H_n=1+\frac{1}{2}+\dots+\frac{1}{n}$. It can be shown that $\ln n<H_n\leq (\ln n)+1$. For $x\geq 1$ the series clearly diverges, so assume $x<1$. Then
$$xn^{\log x}=xe^{\log x\log n}=x^{(\log n)+1}>x^{H_n}>x^{\log n}=e^{\log x\log n}=n^{\log x}.$$
Therefore we can compare with the $p$-series, $p=\log x$. It converges iff $\log x=p<-1$, i.e. $x<\frac{1}{e}$.
(note: $x>0$ is assumed throughout. If $x<0$ elements of the series are not well-defined, $x=0$ is trivial.)
A: It's $\enspace \ln n +\gamma <H_n<\ln(n+1) + \gamma\enspace$ for $\,n\in\mathbb{N}\,$ . 
We get $\enspace\displaystyle\frac{1}{e^\gamma(n+1)}<\frac{1}{e^{H_n}}<\frac{1}{e^\gamma n}\enspace$ and therefore $$ e^{-\gamma x}(\zeta(x)-1)< \sum\limits_{n=1}^\infty e^{-xH_n} <e^{-\gamma x}\zeta(x)$$ 
which is valid for $\,x>1\,$ . 
It follows that $\enspace\displaystyle\sum\limits_{n=1}^\infty x^{H_n}\enspace$ exists for $\enspace\displaystyle 0\leq x<\frac{1}{e}\,$ .
