Question

Use the root test to determine convergence or divergence of the series

E^∞ 100n^2/e^n
n=1

Answer 1

\[\lim_{n \rightarrow \infty}\sqrt[n]{a(n)}=\lim_{n \rightarrow \infty}\sqrt[n]{\frac{ 100n^2 }{e^n }}=\\lim_{n \rightarrow \infty}\sqrt[n]{\frac{ 100n^2 }{e^n }}=\lim_{n \rightarrow \infty}\frac{ \sqrt[n]{100n^2} }{ \sqrt[n]{e^n}}=\lim_{n \rightarrow \infty}\frac{ \sqrt[n]{100} \sqrt[n]{n^2} }{ e}=\frac{ 1 }{ e }<1\]
so it converges