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Is the sum $$\sum_{n=1}^{\infty} \frac{1}{n p_n}$$ where $p_n$ is the n-th prime number converge? I try to check in WolframAlpha and it does not converge, but I don't quite understand the proof due to it's lack of detail. Can someone please provide me with with explanation as to why?

If it does, what number does it converge to?

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    $\begingroup$ Hint: $p_n > n$ – What exactly did you check in WolframAlpha, and what makes you think that the series does not converge? $\endgroup$
    – Martin R
    Nov 2, 2023 at 8:23
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    $\begingroup$ since $p_n\geq n$ for all $n\geq 2$, the series does converges. $\endgroup$
    – Surb
    Nov 2, 2023 at 8:23
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    $\begingroup$ $p_n\sim n \ln n$ $\endgroup$ Nov 2, 2023 at 8:23
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    $\begingroup$ If you're talking about this WA query, I have no idea how it reaches the conclusion that it diverges. It says it uses the integral test (which would not be my first choice due to the discrete nature of the function $n\mapsto p_n$), and then it stops, because I don't have pro access. $\endgroup$
    – Arthur
    Nov 2, 2023 at 8:53
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    $\begingroup$ The value is a bit larger than $0.84897$. $\endgroup$ Nov 2, 2023 at 8:55

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Since not every number is prime, $p_n\ge n$. Therefore, $$\sum_{n=1}^\infty\frac1{np_n}\le\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6$$ Therefore the series does converge, although I am not sure what the series converges to.

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