# Convergence of $\sum_{n=2}^\infty {1\over(\log n)^{\log n}}$.

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.

• Is it $\log(n^{\log n})$ or $\log(n)^{\log(n)}$? – kingW3 Feb 2 '15 at 18:07
• The latter, $(\log n)^{\log n}$. – Tim Raczkowski Feb 2 '15 at 18:07
• Why to avoid the wonderful condensation test? It, together with the $\;n$-th root test, pretty easily tell us there's convergence. – Timbuc Feb 2 '15 at 18:11
• Well, yes you're correct, but Apostol doesn't cover it. So, there must be a way to do the problem without that test. – Tim Raczkowski Feb 2 '15 at 18:14
• You're right that your integral test must be wrong, since it seems to be converging to $\approx 5.7169706$. – JamalS Feb 2 '15 at 18:16

## 2 Answers

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.

• Shouldn't it be $\log\log n>2$ to compare it with $\frac{1}{x^2}$ – avz2611 Feb 3 '15 at 12:40
• @avz2611:$\sum_{n\geq 1}\frac{1}{n^{1.00000001}}$ is converging, too. – Jack D'Aurizio Feb 3 '15 at 13:18
• oh i did not know that , so the exponent if greater than one , series converges? – avz2611 Feb 3 '15 at 13:50
• @avz2611: exactly. – Jack D'Aurizio Feb 3 '15 at 17:01

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.