Originally Posted by

**ruanyueling** My problem is Examine the series $\displaystyle \sum_{n = 2}^\infty \frac{{n}^{\ln n }}{{\ln n}^{ n} }$ . I solved that:

Applying the Cauchy's test we have:

$\displaystyle \sqrt[n]{\frac{{n}^{\ln n }}{{\ln n}^{ n} } } = \frac{{n}^{\frac{\ln n}{ n} } }{\ln n }$

This is wrong: $\displaystyle \sqrt[n]{\ln n^n}=\sqrt[n]{n\ln n}...$

and $\displaystyle \lim_{n \to \infty} \frac{{n}^{\frac{\ln n}{ n} } }{\ln n } = 1$ ( since let $\displaystyle y = {n}^{\frac{\ln n}{ n}$ we have

$\displaystyle \ln y = \frac{{\ln n}^{2 } }{n }= \frac{2}{n }=0$, while $\displaystyle \lim_{n \to \infty}\ln n= infinite => \lim_{n \to \infty}\frac{{n}^{\ln n }}{{\ln n}^{ n} }= 0 <1$.

Hence, the series converges. Is that right?