Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question
    
You've tried a comparison test? –  J. M. Nov 8 '12 at 11:17
2  
The answer seems to follow directly from the equivalent $p_n\sim n\log n$. –  Did Nov 8 '12 at 11:23

3 Answers 3

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.

share|improve this answer
    
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.

share|improve this answer

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

and since

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

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

share|improve this answer
1  
You goofed: $p_n>n.$ –  Charles Nov 8 '12 at 14:37

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.