# The infinite sum $\sum_{m=2}^\infty \space \frac {1} {p_m \space \log\space m}$

Let $p_n$ denote the $n$th prime , for example $p_1$ = $2$ , $p_2 = 3$ etc. Then is the sum $$\sum_{m=2}^\infty \space \frac {1} {p_m \space \log\space m}$$ convergent ?

-
You've tried a comparison test? –  Ｊ. Ｍ. Nov 8 '12 at 11:17
The answer seems to follow directly from the equivalent $p_n\sim n\log n$. –  Did Nov 8 '12 at 11:23

From $$\pi(n)=\frac{n}{\log(n)}+O\left(\frac{n}{\log(n)^2}\right)$$ we get \begin{align} \pi(n\log(n)) &=\frac{n\log(n)}{\log(n\log(n))}+O\left(\frac{n\log(n)}{\log(n\log(n))^2}\right)\\[6pt] &=n\left(1+O\left(\frac{\log(\log(n))}{\log(n)}\right)\right) \end{align} Therefore, $$n=\pi(n\log(n))\left(1+O\left(\frac{\log(\log(n))}{\log(n)}\right)\right)$$ which shows that $$p_n\sim n\log(n)$$ Using this gives that $$\sum_{n=2}^\infty\frac1{p_n\log(n)}$$ converges by comparison to $$\sum_{n=2}^\infty\frac1{n\log(n)^2}$$ which converges by the integral test.

-
This is a better version of my answer, +1. –  Charles Nov 8 '12 at 14:37

This is essentially the same as $$\sum\frac{1}{x\log^2x}$$ which converges since $$\int\frac{dx}{x\log^2x}$$ converges. It's not hard to get an inequality to make this precise.

-
$$\forall\,n\in\Bbb N\,\,\,,\,\,p_n<n\Longrightarrow \frac{1}{p_n\log n}\geq\frac{1}{n\log n}$$
$$\sum_{n=2}^\infty\frac{1}{n\log n}$$
You goofed: $p_n>n.$ –  Charles Nov 8 '12 at 14:37