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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.

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9  
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
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@JoseGarcia Can we consider that elementary? –  Ishan Banerjee Mar 1 '13 at 13:22
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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...") –  anorton Mar 1 '13 at 13:25
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@DavidSpeyer: I'd like to choose your answer if you give me this possibility. :-) –  Chris's sis Mar 1 '13 at 13:30

1 Answer 1

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..

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thanks for your answer (+1) –  Chris's sis Mar 1 '13 at 13:25
    
How we find the asymptotic behavior of $\displaystyle\sum_{j=1}^np_j$? –  Sami Ben Romdhane Mar 1 '13 at 14:37

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