# Determine if $\sum \frac{1}{(\ln{n})^{\ln{n}}}$ converges.

I currently trying to determine it by comparison. I've tried comparing it with $\frac{1}{n^2}$, and it seems to work but I'm not sure if I did it right.

I've done it like this:

$(\ln {n})^{\ln{n}}>n^2$

$e^{\ln{\ln{n}}\times\ln{n}}>e^{\ln{n}\times2}$

$\ln{\ln{n}}\times\ln{n}>\ln{n}\times2$

$\ln{\ln{n}}>2$

$n>e^{(e^2)}$

So I assume this is enough? A convergent series is greater than the original one after some constant $m$. Anything else I need?

• Yes, you are done! – Ma Ming Jun 1 '13 at 21:57
• Well that's fantastic! – Luka Horvat Jun 1 '13 at 21:58
• Yep! And it's about 5.717 for what it's worth. – Sharkos Jun 1 '13 at 22:06
• Is there a way to calculate the actual value or do you just approximate by adding up the first few members? – Luka Horvat Jun 1 '13 at 22:08

$$\frac{1}{(\log n)^{\log n}}=e^{-\log n(\log(\log n))}=\frac{1}{n^{\log(\log n)}}\leq\frac{1}{n^2}$$ for large $n$, hence the given series is convergent by comparison.

• You are making mistake, $e^{(\log n)(\log\log n)}=n^{\log \log n}$. – Ma Ming Jun 1 '13 at 22:02
• You're claiming that $(\log n)^{\log n} = n \log \log n$. Surely that's incorrect. – mrf Jun 1 '13 at 22:03
• It's still incorrect though. – mrf Jun 1 '13 at 22:11
• Finally the conclusion is right. (But the last inequality only holds for sufficiently large $n$.) – mrf Jun 1 '13 at 22:15
• Now it is correct, modulo taking $n$ large enough. But note that it is exactly the OP's argument... – Julien Jun 1 '13 at 22:15

What about the condensation test?

$$2^na_{2^n}=\frac{2^n}{\left(\log2^n\right)^{\log 2^n}}=\frac{2^n}{n^{n\log 2}\left(\log2\right)^{n\log 2}}$$

And now apply the $\,n$-th root test to the above:

$$\sqrt[n]{\frac{2^n}{n^{n\log 2}\left(\log2\right)^{n\log 2}}}=\frac2{n^{log2}\left(\log2\right)^{\log2}}\xrightarrow[n\to\infty]{}0$$

Thus, $\,\sum 2^na_{2^n}\;$ converges $\,\iff\sum a_n\;$ converges.