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Is $\displaystyle\sum\frac1{p^{1+ 1/p}}$ divergent? How can we prove that it is divergent or convergent in analytic number theory? I know what bound of the n-th prime number is, and that its order is $n\log(n)$. Maybe we can use the divergence of $\displaystyle\sum\frac1{n^{1+ 1/n}}$ to show that. I'm not sure that $\displaystyle\sum\frac1{n^{1+ 1/n}}$ is divergent, but I think it is. So would you please help me with this ? Can you help me in finding a proof for it ? Thank you very much, friends.

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  • $\begingroup$ Even if $\sum_{n=1}^{\infty}\frac{1}{n^{1+1/n}}$ diverges, that doesn't tell you much, since all we know is that this sum is larger than the sum with primes. If it converges, however... Also see this guide for how to write math on this site. Welcome to math.stackexchange. $\endgroup$ – Arthur Jan 16 '15 at 10:21
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This is a great question. But you can show that $$\frac 1{p^{1+1/p}} > \frac 1{2p}$$ since $2^p >p$, and thus your series diverges.

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