# Natural logarithm power notation

I am trying to understand how to use Dirichlet's test for convergence and saw an example here (example 2).

Show that $\displaystyle\sum_{i=1}^\infty \frac{2^{2n}n^2}{e^n\,n!}\frac{1}{\ln^2n}$ converges.

Could someone please explain to me what $\ln^2n$ means? Does it mean $\ln(\ln(n))$, $(\ln(n))^2$ or something else entirely?

PS: I'm a newbie in both mathematical analysis and math SE. I spent quite a bit of effort in typing this so I hope that there aren't any glaring mistakes in the formatting. Please let me know if there are any!

• I guess it's $(\ln n)^2$; it's a very ambiguous notation indeed. – egreg Apr 15 '16 at 10:10
• @egreg I'm hoping that having the context of the question will help. Thanks for helping me edit! – jessica Apr 15 '16 at 10:15
• 1. Use Stirling approximation 2. Combine $4^{n}$ and $n^{n}$ 3. Combine $n^{2}$ and $(\ln n)^2$ 4. Find some sequence which larger than that and converges – openspace Apr 15 '16 at 10:24

Without Stirling's formula: With the $n$th term being $A_n,$ show that $A_{n+1}/A_n\to 0$ as $n\to \infty,$ which is more than enough for convergence, as even the weaker result $\exists r\in (0,1)\;\exists m\;(n>m\implies$ $|A_{n+1}|\leq r|A_n|)$ guarantees absolute convergence by comparison with the geometric series $\sum_n r^n.$
1. Using Stirling : $$yoursum = \sum {\frac{4^{n}n^{2}}{n^{n}(\ln n)^{2}}}$$
2. $$\frac{4^n}{n^n} \le \frac{1}{n^2}$$ for some big $n$