## Calculus: Early Transcendentals 8th Edition

$e^{3/5}$
Given: $\Sigma_{n=0}^{\infty}\dfrac{3^{n}}{5^{n}(n!)}$ which can be written as $\Sigma_{n=0}^{\infty}\dfrac{(3/5)^{n}}{n!}$ As we know: $e^{x}=\Sigma_{n=0}^{\infty}\dfrac{x^{n}}{(n!)}$ Then, $e^{3/5}=\Sigma_{n=0}^{\infty}\dfrac{3^{n}}{5^{n}(n!)}$