For what p>0 the series $\sum_{n=1}^{\infty} p^{\sqrt{n}} $ converges?

Question

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 $p^2>p^{\sqrt{5}},p^{\sqrt{6}},p^{\sqrt{7}},p^{\sqrt{8}}$

Answer 1

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.

Answer 2

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.