Does $ \sum^{\infty}_{3} \frac{n^2}{(ln(ln(n)))^{ln(n)}} $ converge?

My initial feeling is no, due to the decreasing gradient of $ln(x)$ so I'd expect the individual terms to 'not tend to 0 fast enough'.

I have tried a few common convergence tests but I haven't spotted the conclusion:

The ratio test shows the individual terms to tend to something which could be one (which I haven't been able to evaluate properly) so it seems inconclusive.

The integral test doesn't seem to generate an easy solution as I don't see how to integrate such a function or compare it to another one.

My only ideas is some elementary solution using comparison test or perhaps the Cauchy condensation test but I don't see how it helps exactly.

Any help is appreciated.


2 Answers 2


$(\ln ( \ln n ) )^{\ln n} = e^{ \ln n \ln (\ln (\ln n))} = n^{\ln(\ln(\ln n))}$.

So, the summand is $n^{2-\ln(\ln(\ln n))}$. There exists a $N$ such that $2 - \ln \ln \ln n < -1.1$ for all $n \geq N$. Then, we compare the sum with terms $n \geq N$ to $\sum_{n \geq N} \frac{1}{n^{1.1}}$ to see that it converges (since the latter is a convergent $p$-series. The first $N$ terms obviously don't affect the convergence.


Using the Cauchy Condensation Test, we get $$ \sum_{n=3}^\infty\frac{n^2}{(\log(\log(n)))^{\log(n)}}\tag{1} $$ converges if $$ \sum_{n=2}^\infty\frac{2^n2^{2n}}{(\log(n\log(2)))^{\,n\log(2)}}\tag{2} $$ converges.

For $n\ge31467207758$ we have $(\log(n\log(2)))^{\log(2)}\ge9$, therefore, $$ \sum_{n=31467207758}^\infty\frac{2^n2^{2n}}{(\log(n\log(2)))^{\,n\log(2)}}\le\sum_{n=31467207758}^\infty\left(\frac89\right)^n\tag{3} $$ Thus, $(2)$ converges and therefore, $(1)$ converges.


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