ln(n)/n^2 Converge or Diverge?

[runningeagle]

Hi, I am looking at

, that's from n=1 to infinity of NATURAL log (n) /n^2.

I am looking to prove whether it converges or diverges.

First I tried the Direct comparison test. I looked at 1/n, but that failed because a[sub:3dxysnt4]n[/sub:3dxysnt4] is less than the divergent 1/n, so inconclusive.

Then I tried the Limit Comparison test. I looked at b[sub:3dxysnt4]n[/sub:3dxysnt4]= 1/n^2, and I get that lim n->inf. a[sub:3dxysnt4]n[/sub:3dxysnt4]/b[sub:3dxysnt4]n[/sub:3dxysnt4]=infinity. Does this mean that ln(n)/n^2 also converges, because 1/n^2 converges?

Thank you.

 

[galactus]

Did you try the integral test?.

\(\displaystyle \int_{1}^{\infty}\frac{ln(n)}{n^{2}}\)

\(\displaystyle \lim_{L\to {\infty}}\int_{1}^{L}\frac{ln(n)}{n^{2}}=\lim_{L\to {\infty}}\left[\frac{-ln(L)}{L}-\frac{1}{L}+1\right]=1\)

So, the integral converges and so does the series. If you are wondering what it converges to, it converges to \(\displaystyle -{\zeta}(1,2)\)=.9375482543

This is related to the Zeta function.