• May 21st 2008, 04:44 PM
szpengchao
• May 21st 2008, 04:45 PM
szpengchao
result
i know this result, alpha <= 1, it is then divergent...but how to prove that?
• May 21st 2008, 04:46 PM
ThePerfectHacker
When $\displaystyle \alpha > 1$, use the integral test.
• May 21st 2008, 04:49 PM
szpengchao
cheers
cheers mate!!!
• May 21st 2008, 05:09 PM
Mathstud28
Quote:
I

$\displaystyle \sum\frac{1}{n\ln(n)}$

$\displaystyle \int_3^{\infty}\frac{dn}{n\ln(n)}=\ln(\ln(n))\bigg |_3^{\infty}=\infty$

divergent

ii

$\displaystyle \sum_{n=3}^{\infty}\frac{1}{n\ln(n)(\ln\ln(n)))^2}$

$\displaystyle \int\frac{dn}{n\ln(n)(\ln\ln(n)))^2}=\frac{-1}{\ln\ln(n))}\bigg|_3^{\infty}=\frac{1}{\ln(\ln(3 ))}$

convergent

iii

$\displaystyle \sum_{n=3}^{\infty}\frac{1}{n^{1+\frac{1}{n}}\ln(n )}$

Limit comparison test with $\displaystyle \frac{1}{n\ln(n)}$

$\displaystyle \lim_{n\to\infty}\frac{\frac{1}{n\ln(n)}}{\frac{1} {n^{1+\frac{1}{n}}\ln(n)}}=\lim_{n\to\infty}n^{\fr ac{1}{n}}\cdot\lim_{n\to\infty}\frac{n\ln(n)}{n\ln (n)}$

Side note $\displaystyle L=\lim_{n\to\infty}n^{\frac{1}{n}}\Rightarrow{ln(L )=\lim_{n\to\infty}\frac{\ln(n)}{n}=0}$

so $\displaystyle \ln(L)=0\Rightarrow{L=e^0=1}$

so $\displaystyle \lim_{n\to\infty}n^{\frac{1}{n}}\cdot\lim_{n\to\in fty}\frac{n\ln(n)}{n\ln(n)}=1\cdot{1}=1$

(EDIT: I should rephrase so you dont misinterpret) Since the limit converges to a finite value and the comparison series diverges, this series diverges