# Convergence of $\sum\limits_{n=3}^\infty \frac1{(\log(\log n))^{\log n}}$

I can prove the sum $\sum\limits_{n=3}^\infty \dfrac1{(\log(\log n))^{\log n}}$ converges this way:

I assume that $\exists n_0 \in \mathbb{N}$ such that $\forall n, n \ge n_0$ we have that $$\frac1{(\log(\log n))^{\log n}} \le \frac1{n^2}.$$

That is $$\log(\log n))^{\log n} \ge n^2.$$

Solving this, I get that $n \ge e^{e^{e^2}}$. From here it follows that the given sum converges, as the sum of the first $n_0 - 1$ terms is finite and the sum converges for $n \ge n_0$ by the comparison test.

This feels extremely awkward.

Is this a correct approach? If there is a better way, could you please show me?

Thank you.

• Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see meta. – Lord_Farin Aug 19 '13 at 9:24
• This is correct--and probably the most direct approach. – Did Aug 19 '13 at 9:24
• @Lord_Farin Thanks for a heads-up. – David Aug 19 '13 at 9:34

Cauchy Condensation says that the given series converges if \begin{align} \sum_{n=2}^\infty\frac{2^n}{\log(n\log(2))^{n\log(2)}} &=\sum_{n=2}^\infty\left(\frac{e}{\log(n\log(2))}\right)^{n\log(2)}\tag{1} \end{align} converges. For $n\ge\frac{e^{2e}}{\log(2)}$, $\frac{e}{\log(n\log(2))}\le\frac12$, and so $(1)$ can be compared to $$\sum_{n=2}^\infty\frac1{2^{n\log(2)}}\tag{2}$$ which is a geometric series.

• Could you please tell me where does the restriction $n\ge\frac{e^{2e}}{\log(2)}$ come from? – David Aug 19 '13 at 9:48
• @DavidČepelík: that ensures that $\frac{e}{\log(n\log(2))}\le\frac12$ – robjohn Aug 19 '13 at 9:50
• Sorry, I am still missing why this is important. By "can be compared", do you mean to take the limit $\lim_{n \to\infty} \frac{\frac{e}{\log{n\log(2)}}}{\frac{1}{2^{n\log(2)}}}$ or to conclude that the limit converges by the comparison test? In the latter case, it would make sense to require $\frac{e}{\log(n\log(2))} \le \frac12$. – David Aug 19 '13 at 11:16
• Use the Comparison Test since $$\left(\frac{e}{\log(n\log(2))}\right)^{n\log(2)}\le\frac1{2^{n\log(2)}}$$ because, as I said in my last comment, when $n\ge\frac{e^{2e}}{\log(2)}$, $$\frac{e}{\log(n\log(2))}\le\frac12$$ – robjohn Aug 19 '13 at 12:31
• Thank you, it is now clear. – David Aug 19 '13 at 12:41

Yes your proof is valid and here is another idea that isn't very different to your work $(\log(\log(n))^{\log n}=n^{\log(\log(\log n))}$ and clearly there's $n_0$ such that $\log(\log(\log n))\geq2$

• It's different in an important sense, namely it doesn't rely on guessing that some condition will hold when you don't know a priori whether it will hold or not. Instead the first step is much more transparent: use a general fact about exponents and logarithms. – Qiaochu Yuan Aug 19 '13 at 9:34
• @QiaochuYuan I said that his work isn't very different because I think that he solved $\log(\log n))^{\log n} \ge n^2$ using the exponents and logarithms. – user63181 Aug 19 '13 at 9:49
• @SamiBenRomdhane: I agree with you. :+) – mrs Nov 1 '13 at 18:12

Use the fact that $a^b = e^{b \log a}$ twice to write $(\log \log n)^{\log n} = e^{\log n \log \log \log n} = n^{\log \log \log n}$. This will be greater than $n^2$ once $\log \log \log n \ge 2$, or once $n \ge e^{e^{e^2}}$ as you observed.