$\displaystyle \sum_{n=1}^{\infty}\ \frac{k^2}{k^3+1}$
Follow Math Help Forum on Facebook and Google+
Use limit comparsion using $\displaystyle \frac{1}{n}$.
i did and$\displaystyle \frac{k^2}{k^3+1}< \frac{1}{k}$though ,so its inconclusive
nevermind i used the limit comparison test and got the answer thank you
Do you know the limit comparsion test? $\displaystyle a_n=\frac{n^2}{n^3+1}~\&~b_n=\frac{1}{n}$.
View Tag Cloud