Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How would we test for convergence the series below?

$$\sum_{j=1}^{\infty}\frac{1}{\sum_{i=1}^{j}p_i}$$ where $p_i$ is the $i$th prime number. I'd be glad to learn an elementary way. Thanks.

share|cite|improve this question
Hint: $p_j \geq j$. – David Speyer Mar 1 '13 at 13:21
the sum is convergent in think :D , by prime number theorem – Jose Garcia Mar 1 '13 at 13:21
@JoseGarcia Can we consider that elementary? – Ishan Banerjee Mar 1 '13 at 13:22
Where do you get all these crazy questions? (Every time I see a question by "Chris's sister and pals", I think to myself "This is going to be an interesting/hard question...") – apnorton Mar 1 '13 at 13:25
@DavidSpeyer: I'd like to choose your answer if you give me this possibility. :-) – user 1618033 Mar 1 '13 at 13:30
up vote 6 down vote accepted

by priem number theorem $$ \sum_{j=1}^{n}p_{j} \sim \frac{n^{2}ln(n)}{2} $$

so your series goes about $$ \sum_{n=2}^{\infty} \frac{2}{n^{2}ln(n)} $$

which is itself convergent..

share|cite|improve this answer
thanks for your answer (+1) – user 1618033 Mar 1 '13 at 13:25
How we find the asymptotic behavior of $\displaystyle\sum_{j=1}^np_j$? – user63181 Mar 1 '13 at 14:37

Your Answer


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.