Is the series convergent $x + x^{1+\frac{1}{2}}+x^{1+\frac{1}{2}+ \frac{1}{3}}+\cdots$

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).

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.)