I am interested in finding a tight upper bound for the following summation, for values of $\alpha > 0$ (but for my particular application, I would be completely satisfied just to understand the cases $0 < \alpha \leq 1/2$):
$$f(x) = \sum_{p=2}^\infty \left(\frac{x}{p^\alpha}\right)^p$$
Here's what I've tried. In the case that $\alpha = 0$, the $p^\alpha$ in the denominator is $1$ and then this becomes the geometric series, and $f(x)$ and its radius of convergence is completely understood. When $\alpha > 0$ this series converges for all $x$ because for large enough $p$ the fraction is $<1$, and then all remaining terms can be bounded by a convergent geometric series. Specifically, letting $p^*$ be the smallest $p$ such that $\frac{x}{p^\alpha} \leq 1/2$, then $$f(x) \leq \sum_{p=2}^{p^*-1}\left(\frac{x}{p^\alpha}\right)^p + \sum_{p = p^*}^\infty(1/2)^p = \sum_{p=2}^{p^*-1}\left(\frac{x}{p^\alpha}\right)^p + \left(\frac{1}{2}\right)^{p^*}$$
I still don't know how to deal with the first term, or if this is a useful start. I would greatly appreciate more ideas. Thanks!
I don't know why I thought that the question was on convergence radius. My bad. I'll leave the answer in case someone finds it useful.
I will use $n$ instead of $p$ to index the summation. Let $a= e^\alpha$. Rewrite the series as:
$$ \sum_{n=2}^\infty \left( \frac{x}{n^\alpha}\right)^n = \sum_{n=2}^\infty \frac{x^n}{n^{n\alpha}} = \sum_{n=2}^\infty \frac{x^n}{a^{n\log n}} $$
By the root test $$ \limsup_{n\to\infty}\sqrt[n]{\frac{|x^n|}{|a^{n\log n}|}} = \limsup_{n\to\infty}\frac{|x|}{a^{\log n}} =\begin{cases} |x| &\text{ if $a = 1 \iff \alpha = 0$ } \\ 0 &\text{ if $a > 1 \iff \alpha > 0$ } \\ \infty &\text{ if $a < 1 \iff \alpha < 0$ } \end{cases} $$
Therefore: