For what p>0 the series $\sum_{n=1}^{\infty} p^{\sqrt{n}} $ converges? My opinion is that it converges for all $0<p<1$
For $p\geq 1$, $p^{\sqrt{n}}$ does not converge to zero so it cannot converge.
Now, for $0<p<1$ I tried to the following:
$\sum_{n=1}^{\infty} p^{\sqrt{n}} =p+p^{\sqrt{2}}+p^{\sqrt{3}}+p^2+...\leq 3p+5p^2+7p^3+...=\sum_{n=1}^{\infty} (2n+1)p^{n}$
Using the quotient test, $\frac{(2n+3)p^{n+1}}{(2n+1)p^{n}}$ converges to $p$ and $p<1$, so the series $\sum_{n=1}^{\infty} (2n+1)p^{n}$ converges and therefore  $\sum_{n=1}^{\infty} p^{\sqrt{n}} $ as well.
Is this valid? 
EDIT: I got $p+p^{\sqrt{2}}+p^{\sqrt{3}}+p^2+...\leq 3p+5p^2+7p^3+...$ by using $p>p^{\sqrt{2}},p^{\sqrt{3}}$, and then $p^2>p^{\sqrt{5}},p^{\sqrt{6}},p^{\sqrt{7}},p^{\sqrt{8}}$
 A: You should add some words to make it more clear. Or at least add parentheses to show how you are grouping terms. It took me a second to see that you were doing something like this:
\begin{align}\sum_{n=1}^\infty p^{\sqrt n}&=p+p^{\sqrt2}+p^{\sqrt3}\tag{$<3p$}\\&\hphantom{~\!=}+p^{\sqrt4}+p^{\sqrt5}+p^{\sqrt6}+p^{\sqrt7}+p^{\sqrt8}\tag{$<5p^2$}\\&\hphantom{~\!=}+p^{\sqrt9}+p^{\sqrt{10}}+p^{\sqrt{11}}+p^{\sqrt{12}}+p^{\sqrt{13}}+p^{\sqrt{14}}+p^{\sqrt{15}}\tag{$<7p^3$}\\&\hphantom{~\!=}+\dots\\&<3p\\&\hphantom{~\!=}+5p^2\\&\hphantom{~\!=}+7p^3\\&\hphantom{~\!=}+\dots\\&=\sum_{n=1}^\infty(2n+1)p^n\end{align}
So indeed your proof is correct, but needs clarification.
A: Note:
I added a proof that
the series converges for
$\sum p^{n^a}$
for any $0 < a$.
If
$0 < p < 1$,
then
$p = \dfrac1{1+q}$
where $q > 0$.
Then
$\begin{array}\\
p^{\sqrt{n}}
&=\dfrac1{(1+q)^{\sqrt{n}}}\\
&=\dfrac1{\exp(\ln(1+q)\sqrt{n})}\\
&<\dfrac{4!}{(\ln(1+q)\sqrt{n})^4}
\qquad\text{since }\exp(x) > x^4/4!\\
&=\dfrac{24}{n^2\ln^4(1+q)}\\
&=\dfrac{24}{n^2\ln^4(p)}
\qquad\text{since }\ln(1+q) = -\ln(p)\\
\end{array}
$
and the sum of these converges.
The inequality
$\exp(x) > x^4/4!$
was moderately arbitrarily chosen from
$\exp(x) > x^m/m!$;
any even $m> 2$
would have worked.
Note that this works
for $n^a$
for any $a > 0$:
$\begin{array}\\
p^{n^a}
&=\dfrac1{(1+q)^{n^a}}\\
&=\dfrac1{\exp(\ln(1+q)n^a)}\\
&<\dfrac{(2m)!}{(\ln(1+q)n^a)^{2m}}
\qquad\text{since }\exp(x) > x^{2m}/(2m)!\\
&=\dfrac{(2m)!}{n^{2ma}\ln^{2m}(1+q)}\\
&=\dfrac{(2m)!}{n^{2ma}\ln^{2m}(p)}
\qquad\text{since }\ln(1+q) = -\ln(p)\\
\end{array}
$
If we choose
$m$ so that
$ma > 1$,
then each term is less than
$\dfrac{(2m)!}{n^{2ma}\ln^{2m}(p)}
\lt \dfrac{(2m)!}{n^{2}\ln^{2m}(p)}
$
and the sum of these converges.
