$\displaystyle \sum_{k=2}^\infty ln\frac{k^2-1}{k^2}$ I'm trying to figure out how to find the limit of this series. Any advice?
Follow Math Help Forum on Facebook and Google+
Originally Posted by paupsers $\displaystyle \sum_{k=2}^\infty ln\frac{k^2-1}{k^2}$ I'm trying to figure out how to find the limit of this series. Any advice? $\displaystyle \frac{k^2-1}{k^2}=\frac{k-1}{k} \,\ \frac{k+1}{k}$. Apply: $\displaystyle ln(ab)=ln(a)+ln(b)$. for $\displaystyle a,b>0$.
View Tag Cloud