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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 ?

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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
up vote 2 down vote accepted

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.

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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.

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$$\forall\,n\in\Bbb N\,\,\,,\,\,p_n<n\Longrightarrow \frac{1}{p_n\log n}\geq\frac{1}{n\log n}$$

and since

$$\sum_{n=2}^\infty\frac{1}{n\log n}$$

diverges (for example, using Cauchy's Condensation Test), our series also diverges.

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You goofed: $p_n>n.$ – Charles Nov 8 '12 at 14:37

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