Determine if the series $$ \sum_{n=2}^{\infty} \frac{1}{( \ln \ln n)^{\ln \ln n}} $$ converges.
The terms of the sequence do not go to zero and the ratio test gives no useful information, so I am trying to find a series to compare it with. I am pretty sure that the series diverges, so I'm trying to compare it with $1/n$ but am getting stuck.
On
This is a typical case for Cauchy condensation test.
Let replace $a_n=\dfrac 1{\ln\ln(n)^{\ln\ln(n)}}$ by $b_n=2^na_{2^n}$ to get the condensed series
$b_n=\dfrac {2^n}{\ln(n\ln(2))^{\ln(n\ln(2))}}$
And let's do it again
$c_n=\dfrac{2^n2^{2^n}}{(\ln\ln(2)+n\ln(2))^{(\ln\ln(2)+n\ln(2))}}$
The denominator in $(a+bn)^{a+bn}$ is roughly in $(bn)^{bn}$ which is negligible compared to $2^{2^n}$ on the numerator, so $c_n\not\to 0$.
So the series $\sum c_n$ is divergent as well as $\sum b_n$ and eventually $\sum a_n$.
Note: if we could do it with $e$ instead of $2$, the condensation test would lead immediately to a more visual result, indeed we would get:
$c_n=\dfrac{e^ne^{e^n}}{n^n}=\exp(n+e^n-n\ln(n))\sim \exp(e^n)\to\infty$. Unfortunately the test requires integer indices, and we have to deal with these annoying $\ln(2)$ terms. But doing so gives a good indication whether the series converges or not.
Let's look at the general term of the series in a less confusing form. For any (sufficient big, so that the quantity $\ln\ln n$ is defined) integer $n$, $$ (\ln\ln n)^{\ln\ln n} = e^{\ln\ln n\cdot \ln\ln\ln n}\,. $$ Now, we have $$ n = e^{\ln n} $$ and since $$ \ln n > (\ln \ln n)^2 > \ln\ln n\cdot \ln\ln\ln n $$ for $n$ big enough (and $\exp$ is increasing), it follows that $$ n > (\ln\ln n)^{\ln\ln n} $$ for $n$ big enough, and therefore (again for $n$ big enough) $$ \frac{1}{n} < \frac{1}{(\ln\ln n)^{\ln\ln n}}\,. $$ You can now conclude by comparison to $\sum_n \frac{1}{n}$.