Answer
The series is convergent and the sum of the series is $e-1$.
Work Step by Step
Define the $n-$th partial sum: $s_n=\sum_{i=1}^n(e^{1/i}-e^{1/(i+1)})$.
Simplify $s_n$:
$s_n=\sum_{i=1}^n(e^{1/i}-e^{1/(i+1)})=e^{1/1}-e^{1/(n+1)}=e-e^{1/(n+1)}$
Find the sum of the series:
$\sum_{n=1}^\infty(e^{1/n}-e^{1/(n+1)})=\lim\limits_{n \to \infty}
s_n=\lim\limits_{n \to \infty}(e-e^{1/(n+1)})=e-e^{\lim\limits_{n \to \infty}1/(n+1)}=e-e^0=e-1$
So, the series is convergent and the sum of the series is $e-1$.
