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I've been working through exercises in Chapter 8 of Apostol's Mathematical Analysis. Exercise 8.15 gives a number series to be tested for convergence. I've gotten most of them but I'm stuck on

$$\sum_{n=1}^\infty{1\over(\log n)^{\log n}}.$$

The root test is in conclusive, and the things I can think of to compare with have the wrong inequality. I thought I had proven divergence by the integral test, but something was wrong because according to Wolfram Alpha, this converges. Hint, anyone?

P.S. Looking to avoid Cauchy condensation test.

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  • $\begingroup$ Is it $\log(n^{\log n})$ or $\log(n)^{\log(n)}$? $\endgroup$
    – kingW3
    Feb 2, 2015 at 18:07
  • $\begingroup$ The latter, $(\log n)^{\log n}$. $\endgroup$
    – Tim Raczkowski
    Feb 2, 2015 at 18:07
  • $\begingroup$ Why to avoid the wonderful condensation test? It, together with the $\;n$-th root test, pretty easily tell us there's convergence. $\endgroup$
    – Timbuc
    Feb 2, 2015 at 18:11
  • $\begingroup$ Well, yes you're correct, but Apostol doesn't cover it. So, there must be a way to do the problem without that test. $\endgroup$
    – Tim Raczkowski
    Feb 2, 2015 at 18:14
  • $\begingroup$ You're right that your integral test must be wrong, since it seems to be converging to $\approx 5.7169706$. $\endgroup$
    – JPhy
    Feb 2, 2015 at 18:16

2 Answers 2

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Hint: $$(\log n)^{\log n} = n^{\log\log n} $$ and as soon as $\log\log n > 1$, you can compare your series with a converging one.

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  • $\begingroup$ Shouldn't it be $\log\log n>2$ to compare it with $\frac{1}{x^2}$ $\endgroup$
    – avz2611
    Feb 3, 2015 at 12:40
  • $\begingroup$ @avz2611:$\sum_{n\geq 1}\frac{1}{n^{1.00000001}}$ is converging, too. $\endgroup$
    – Jack D'Aurizio
    Feb 3, 2015 at 13:18
  • $\begingroup$ oh i did not know that , so the exponent if greater than one , series converges? $\endgroup$
    – avz2611
    Feb 3, 2015 at 13:50
  • $\begingroup$ @avz2611: exactly. $\endgroup$
    – Jack D'Aurizio
    Feb 3, 2015 at 17:01
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Hint: Cauchy condensation Test

$\sum u(n)$ and $\sum u(a^n) a^n$ converge/diverge together for $a>0$.

Hence, test $\sum \dfrac {1}{(\log a^{n})^{ \log a^n }} \cdot a^n$ for convergence.

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