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I found this series but in the convergence proof I don't know if it is correct to say that it converges.

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  • $\begingroup$ What convergence tests have you tried to apply? $\endgroup$
    – Ramanujan
    Sep 22 at 17:59
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    $\begingroup$ This question isn’t at all clear to me. If you’ve seen a proof that it converges, then the problem is…? $\endgroup$
    – FShrike
    Sep 22 at 17:59
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    $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Sep 22 at 18:01
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    $\begingroup$ Show how you concluded the convergence , otherwise it is impossible for us to tell you whether your argument is valid. $\endgroup$
    – Peter
    Sep 22 at 18:03
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    $\begingroup$ But, hint: it can be easier to consider $\log(n)/n$ in terms of $x/e^x$ $\endgroup$
    – FShrike
    Sep 22 at 18:07

1 Answer 1

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I don't see any problem here. Just apply the root test:

If $\displaystyle \lim_{n}\sqrt[n]{a_{n}}=r<1 \,\,\,$then the series converges.

But $\displaystyle \lim_{n}\sqrt[n]{(\dfrac{logn}{n}})^{n}$=$\displaystyle \lim_{n}\dfrac{logn}{n}= \displaystyle \lim_{n}log\sqrt[n]{n}=log1=0<1$ . Thus the series converges.

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