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For what values of $p$ does the infinite sum of $p^{\sqrt{n}}$ over $n$ from $1$ to infinity converge?

I thought to consider it in relation to a geometric series, but found that $\lim_{n \rightarrow \infty} \frac{p^{\sqrt{n}}}{p^n} = \infty$. So now, I am not sure how to think about this. By graphing it for some values of $p$, it seems to converge for $p \in [0,1]$.

algebroo
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1 Answers1

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I took a look at the solutions here

None of them makes use of the generalized Cauchy condensation test. In the OP case for $0<p<1$ the sequence $p^{\sqrt{n}}$ is decreasing. Therefore we may apply the condensation test. Choose $n_k=k^2.$ The new series becomes $$\sum (2k+1)p^k $$ which is convergent in an obvious way, say by the ratio test, or by applying the Abel transformation (summation by parts).

Ryszard Szwarc
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    How can you choose $n_k=k^2$? Which form of the Cauchy condensation test are you using? – Mark Apr 05 '23 at 15:27
  • If $a_n\searrow 0$ and $n_k$ satisfies $$d(n_k-n_{k-1})\le n_{k+1}-n_k\le c (n_k-n_{k-1})$$ for some positive constants $c$ and $d$ then $$\sum a_n<\infty \iff \sum (n_{k+1}-n_k)a_{n_k}<\infty$$ The most popularchoice is $n_k=2^k,$ but sometimes other choices are more convenient. – Ryszard Szwarc Apr 05 '23 at 16:17
  • That's interesting. I know you can replace $2^k$ with $p^k$ for some $p>1$, but never heard about that generalization. – Mark Apr 05 '23 at 16:23
  • @Mark The proof goes along the same lines as that for $n_k=2^k.$ Each direction makes use of one of the inequalities. In the case of $n_k=2^k$ the ratio $(n_{k+1}-n_k)/(n_k-n_{k-1})$ is constant and equal $2.$ see – Ryszard Szwarc Apr 05 '23 at 16:27