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I came across a problem on convergence of series and I did not get into any idea about this -- any help or hints ? $$\sum_{n=2}^\infty {1\over n^{\log n}}$$

What about $n^{\log(\log n)}$ ?

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  • $\begingroup$ Making a guess as to what you mean, Baby Rudin 3.28-3.29 would seem helpful. $\endgroup$
    – ForgotALot
    Mar 7, 2016 at 5:59
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    $\begingroup$ To show divergence if $a\le 1$, it is enough (why?) to show divergence when $n=1$. For this use the Integral Test. $\endgroup$ Mar 7, 2016 at 6:11
  • $\begingroup$ Hello can you make it clear that my question is about n power logn $\endgroup$ Mar 7, 2016 at 9:20
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    $\begingroup$ For some $N$, $n>N\implies\log(n)>1$. For some other $N$, $n>N\implies\log(\log(n))>1$. $\endgroup$
    – user65203
    Mar 7, 2016 at 9:24
  • $\begingroup$ Can you edit the problem with n power to the logn please i am not able to do so $\endgroup$ Mar 7, 2016 at 9:25

2 Answers 2

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For which $a$ does $\sum_{n=2}^\infty {1\over n^a} $ converge? For which $a$ does it diverge?

If this converges, then $\sum_{n=2}^\infty {1\over n^a\log n}$ will certainly converge (do you see why?).

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Since the series is a positive one and the general term's sequence is monotonically decreasing to zero, you can use Cauchy's Condensation Test, and:

$$\frac{2^n}{(2^n)^{\log2^n}}=\frac{2^n}{2^{n^2\log2}}=\frac1{2^{n\left(n\log2-1\right)}}\le\frac1{2^n}$$

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  • $\begingroup$ Aaaa that is there . Thanks . $\endgroup$ Mar 7, 2016 at 10:16
  • $\begingroup$ Thank you to anyone who downvoted/upvoted this. It would really help me if anyone who downvoted could be please kind enough to write down the reason: is something wrong, or unclear? I try to learn of my mistakes...if I recognize them. $\endgroup$
    – DonAntonio
    Mar 7, 2016 at 10:32

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