## Calculus: Early Transcendentals 8th Edition

$\Sigma_{n=1}^{\infty}a_{n}=\Sigma_{n=1}^{\infty}\frac{e^{1/n} }{n^{2}}$ We will compare with the series $\Sigma_{n=1}^{\infty}b_{n}=\Sigma_{n=1}^{\infty}\frac{e}{n^{2}}$ we have: $a_{n}=\frac{e^{1/n} }{n^{2}} \leq \frac{e}{n^{2}}=b_{n}$ This is because $e^{1/n}\leq e$ for all $n\geq 1$ Thus, $\Sigma_{n=1}^{\infty}b_{n}=\Sigma_{n=1}^{\infty}\frac{e}{n^{2}}=e \cdot \Sigma_{n=1}^{\infty}\frac{1}{n^{2}}$ As a series $\Sigma_{n=1}^{\infty}\frac{1}{n^{2}}$converges because it is a p-series with $p=2$. Hence, the given series converges by the comparison test.