# Help finding the limit of this series

• Feb 15th 2010, 03:57 PM
paupsers
Help finding the limit of this series
$\sum_{k=2}^\infty ln\frac{k^2-1}{k^2}$

I'm trying to figure out how to find the limit of this series. Any advice?
• Feb 15th 2010, 04:30 PM
General
Quote:

Originally Posted by paupsers
$\sum_{k=2}^\infty ln\frac{k^2-1}{k^2}$

I'm trying to figure out how to find the limit of this series. Any advice?

$\frac{k^2-1}{k^2}=\frac{k-1}{k} \,\ \frac{k+1}{k}$.

Apply:

$ln(ab)=ln(a)+ln(b)$. for $a,b>0$.