# Convergence of $\sum_{n=1}^{\infty}(\frac{H_n}{p_n}-\frac{n}{n^n})$

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.

• I believe that $\sum\dfrac{H_n}{p_n}$ diverges since the sequence of partial sums is increasing. – user122283 Apr 27 '14 at 16:01
• That would mean that $H_n$ increases faster than $p_n$ interesting result! – user146159 Apr 27 '14 at 16:04

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