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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?2012-11-08
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    The answer seems to follow directly from the equivalent $p_n\sim n\log n$.2012-11-08

3 Answers 3

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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, +1.2012-11-08
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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

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.$2012-11-08