# Need a help convergence!

• May 23rd 2011, 04:42 AM
ruanyueling
Need a help convergence!
My problem is Examine the series \sum_{n = 2}^\infty \frac{{n}^{lnn }}{{lnn}^{ n} } . I solved that:
Applying the Cauchy's test we have:
\sqrt[n]{\frac{{n}^{lnn }}{{lnn}^{ n} } } = \frac{{n}^{\frac{lnn}{ n} } }{lnn } and \lim_{n \to \infty} \frac{{n}^{\frac{lnn}{ n} } }{lnn } = 1 ( since let y = {n}^{\frac{lnn}{ n} we have lny = \frac{{lnn}^{2 } }{n }= \frac{2}{n }=0 while \lim_{n \to \infty}lnn= infinite => \lim_{n \to \infty}\frac{{n}^{lnn }}{{lnn}^{ n} }= 0 <1. Hence, the series converges. Is that right?
• May 23rd 2011, 06:03 AM
tonio
Quote:

Originally Posted by ruanyueling
My problem is Examine the series $\sum_{n = 2}^\infty \frac{{n}^{\ln n }}{{\ln n}^{ n} }$ . I solved that:
Applying the Cauchy's test we have:

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

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

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

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

Hence, the series converges. Is that right?

.

Comparison seems to be way easier: $\frac{n^{\ln n}}{n\ln n}\geq \frac{1}{\ln n}$ , and as the series of the rightmost expression diverges so does our series.

Tonio
• May 23rd 2011, 06:18 AM
ruanyueling
ah, sorry. this is (lnn)^n, not lnn^n. ^^