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  ?

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.
