Question

Does the sum from n=2 to infinity of ((ln n)^2n)/(n^n) converge or diverge?

Answer 1

Try using the root test:
\[\lim_{n \to \infty}\left| a_n\right|^{\frac{1}{n}} \]
If the limit \(<1\) the series converges absolutely (and is thus convergent).
If the limit \(>1\), the series diverges.
If the limit \(=1\), inconclusive (Although I did suggest the root test for a reason, so it should be one of the 2 options above ;) )

\[\lim_{n \to \infty} \left|\frac{(\ln n)^{2n}}{n^n} \right|^{\frac{1}{n}} \]. You can drop the absolute value bars since the argument will always be positive (since \(n=2 > 0\) and clearly \(\ln n >0\) and \(n^n>0\) for n=2,3,...
\[\lim_{n \to \infty} \left|\frac{(\ln n)^{2n}}{n^n} \right|^{\frac{1}{n}} =\lim_{n \to \infty}\left[\left(\frac{(\ln n)^2}{n}\right)^n\right]^{\frac{1}{n}}=\lim_{n\to \infty}\frac{(\ln n)^2}{n} \]
Notice this gives the form \(\dfrac{\infty}{\infty}\) so you can use l'Hôpital's rule. The limit should be easy then.