## University Calculus: Early Transcendentals (3rd Edition)

Consider $a_n=\Sigma_{n=1}^\infty \dfrac{1}{1^2+2^2+....n^2}=\Sigma_{n=1}^\infty \dfrac{6}{n(n+1)(2n+1)}$ and $b_n=\dfrac{1}{ n^3}$ Now, $\lim\limits_{n \to \infty}\dfrac{a_n}{b_n} =\lim\limits_{n \to \infty}\dfrac{\dfrac{6}{n(n+1)(2n+1)}}{\dfrac{1}{ n^3}}=\lim\limits_{n \to \infty}\dfrac{6 n^2}{(n+1)(2n+1)}$ or, $=\lim\limits_{n \to \infty} \dfrac{6}{(1+1/n)(2+1/n)}$ and $\lim\limits_{n \to \infty} \dfrac{6}{(1+0)(2+0)}=3$ Here, $\Sigma_{n=1}^\infty \dfrac{1}{n^3}$ is a convergent p-series test with $p \gt 1$ Thus the series converges by the limit comparison test.