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Does this diverge or converge ?? $$\sum_{n=1}^{\infty}(\frac{H_n}{p_n}-\frac{n}{n^n})$$ where $H_n$ is the nth harmonic number, $p_n$ is the nth prime.

My impression is that it diverges, but I don't see how I can prove it! I tried on wolframalpha but no clue.

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  • $\begingroup$ I believe that $\sum\dfrac{H_n}{p_n}$ diverges since the sequence of partial sums is increasing. $\endgroup$ – user122283 Apr 27 '14 at 16:01
  • $\begingroup$ That would mean that $H_n$ increases faster than $p_n $ interesting result! $\endgroup$ – user146159 Apr 27 '14 at 16:04
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A classic result is $$H_n\sim_\infty \ln n$$ and by the Flegner's result$^{(1)}$ in $1990$ we have $$ 0.91\; n \ln(n) < p_n < 1.7\; n \ln(n)$$ hence the series $$\sum_{n=1}^\infty \frac{H_n}{p_n}$$ is divergent. Since the series $$\sum_{n=1}^\infty\frac{n}{n^n}$$ is obviously convergent then given series is divergent.

$(1)$ The page is in French language.

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Hint: $\sum_{n=1}^{\infty } \frac{n}{n^n}<\infty$ but $\sum_{n=1}^{\infty } \frac{1}{p_n}=\infty$.

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