Answer
The series is convergent and the sum is $e-1$.
Work Step by Step
$\Sigma^{\infty}_{n=1} (e^{\frac{1}{n}}-e^{\frac{1}{n+1}})$
$S_{n}=\Sigma^{n}_{k=1} (e^{\frac{1}{k}}-e^{\frac{1}{k+1}})$
$=e^{1} - e^{\frac{1}{2}} + e^{\frac{1}{2}} - e^{\frac{1}{3}} + e^{\frac{1}{3}} - e^{\frac{1}{4}}+...+e^{\frac{1}{n}}-e^{\frac{1}{n+1}}$
$=e^{1}-e^{\frac{1}{n+1}}$
$\lim\limits_{n \to \infty}S_{n} = \lim\limits_{n \to \infty}(e^{1}-e^{\frac{1}{n+1}})$
$=e-e^{\frac{1}{\infty}}$
$=e-e^{0}$
$=e-1$
The series is convergent and the sum is $e-1$.
