Does the series $\sum_{n=1}^{\infty}|x|^\sqrt n$ converge pointwise? If it then what would be the sum? Does the series $\sum_{n=1}^{\infty}|x|^\sqrt n$ converge pointwise?
 A: For $|x|\geq 1$, it diverges as the general term does not tend to $0$.
When $0<|x|<1$, we have
$$
n^2|x|^\sqrt{n}=\exp\left(2\log n +\log|x|\sqrt{n}\right)=\exp\left(\log|x|\sqrt{n}\left(1+\frac{2\log n}{\log|x|\sqrt{n}}\right)\right)\longrightarrow 0.
$$
Since a converging sequence is bounded, it follows that there exists a constant $C>0$ such that
$$
0\leq |x|^\sqrt{n}\leq\frac{C}{n^2}\qquad\forall n\geq 1.
$$
It follows that the series converges absolutely by comparison. 
For $x=0$, it converges to $0$, if we agree that $0^\sqrt{n}=0$ for $n\geq 1$.
Note: I did not see you were also asking about the sum in the title. Unfortunately, I don't know about that and Wolfram Alpha does not give a closed form either in the case $x=1/2$.
A: If $|x|\geqslant1$ then $|x|^\sqrt n\geqslant1$ does not converge to zero hence the series $\sum\limits_{n\geqslant1}|x|^\sqrt{n}$ diverges. If $|x|\lt1$ then for every $n$ in the slab $k^2\leqslant n\lt(k+1)^2$, $|x|^\sqrt{n}\leqslant|x|^k$ hence $\sum\limits_{n\geqslant1}|x|^\sqrt{n}\leqslant\sum\limits_{k\geqslant1}(2k+1)|x|^k$, which proves that the series $\sum\limits_{n\geqslant1}|x|^\sqrt{n}$ converges. 
As Harald Hanche-Olsen mentioned in a comment, this can be viewed as a case of Cauchy's condensation test. 

The sum $S(x)=\sum\limits_{n\geqslant1}|x|^\sqrt{n}$ has no simple expression when $|x|\lt1$. However, $\sum\limits_{k\geqslant1}r^k=\frac{r}{1-r}$ and $\sum\limits_{k\geqslant1}kr^k=\frac{r}{(1-r)^2}$ for every $|r|\lt1$ hence the upper bound explained above shows that $S(x)\leqslant|x|\frac{3-|x|}{(1-|x|)^2}\leqslant\frac{2}{(1-|x|)^2}$. 
Likewise, for every $k^2\leqslant n\lt(k+1)^2$, $|x|^\sqrt{n}\geqslant|x|^{k+1}$ hence $S(x)\geqslant\sum\limits_{k\geqslant1}(2k+1)|x|^{k+1}=|x|^2\frac{3-|x|}{(1-|x|)^2}$. 
In particular, $S(x)\sim\frac2{(1-|x|)^2}$ when $|x|\to1$.
