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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!

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    $\begingroup$ I guess it's $(\ln n)^2$; it's a very ambiguous notation indeed. $\endgroup$ – egreg Apr 15 '16 at 10:10
  • $\begingroup$ @egreg I'm hoping that having the context of the question will help. Thanks for helping me edit! $\endgroup$ – jessica Apr 15 '16 at 10:15
  • $\begingroup$ 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 $\endgroup$ – openspace Apr 15 '16 at 10:24
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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 $\sum_n r^n.$

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Hint:

  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 $n$

Now find upper bound of second part.

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