Show that $\sum_{n=3}^{\infty} \frac{1}{(\log \log n)^{\log \log n}}\quad $ diverges.

Question

This problem comes from Apostol's Ex. 8.15

Show that $\sum_{n=3}^{\infty} \frac{1}{(\log \log n)^{\log \log n}} \;$ diverges.

My attempt:

Note $ \log \log n$ is increasing and positive, so the term $a_n$ is decreasing and positive. Clearly, $\lim a_n = 0$. Thus, we can use the integral test. $$ \int^\infty_3 \frac{dx}{(\log \log x)^{\log \log x} } = \int^\infty_{\log\log 3} \frac{e^{e^u+1}du}{u^u} = \int^\infty_{\log\log 3} e^{e^u+1- u\log u} du. $$ Since $\lim (e^u +1 - u\log u) = \infty$, the integral clearly doesn't converge, thus the series diverges by the integral test.

Is my solution correct? Besides, I'd like to know other approaches different from the integral test. Thanks.

Answer 1

Note that

\begin{align*} \lim_{n \to \infty} \frac{\frac{1}{\log(\log(n))^{\log(\log(n))}}}{\frac{1}{n}} =&\lim_{n \to \infty} \frac{n}{\log(\log(n))^{\log(\log(n))}}\\ =&\lim_{u \to \infty} \frac{e^{e^u}}{u^u}\\ =&\exp\left(\lim_{u \to \infty} (e^u-u \ln u)\right)\\ =&+\infty. \end{align*}

because if $\ln(x) < x$, then $e^x - x \ln x > e^x - x^2,$ and evidently $\lim_{x \to \infty} e^x-x^2 = +\infty.$ So by comparison with harmonic series, your series diverges.