## Calculus 8th Edition

In the given problem $a_{n}=\frac{n^{2n}}{(1+2n^{2})^{2}}$ $\frac{n^{2n}}{(1+2n^{2})^{2}}=(\frac{n^{2}}{(1+2n^{2}})^{2}$ Therefore, $\lim\limits_{n \to \infty}|a_{n}|^{1/n}=\lim\limits_{n \to \infty}\frac{n^{2}}{1+2n^{2}}$ Divide numerator and denominator by $n^{2}$ $=\lim\limits_{n \to \infty}\frac{1}{\frac{1}{n^{2}}+2}$ $=\frac{1}{0+2}$ $=\frac{1}{2} \lt 1$ Hence, the series converges by Root Test.