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 ?
The infinite sum $ \sum_{m=2}^\infty \space \frac {1} {p_m \space \log\space m} $
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2The answer seems to follow directly from the equivalent $p_n\sim n\log n$. – 2012-11-08
3 Answers
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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0This is a better version of my answer, +1. – 2012-11-08
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
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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1You goofed: p_n>n. – 2012-11-08