Answer
The series is convergent and the sum is $\frac{e}{e-1}$.
Work Step by Step
$\Sigma^{\infty}_{n=1} (\frac{1}{e^{n}}+\frac{1}{n(n+1)})$
$=\Sigma^{\infty}_{n=1} \frac{1}{e^{n}} + \Sigma^{\infty}_{n=1}\frac{1}{n(n+1)}$
$=\Sigma^{\infty}_{n=1} (\frac{1}{e})^{n} + \Sigma^{\infty}_{n=1}\frac{1}{n(n+1)}$
First
$\Sigma^{\infty}_{n=1} (\frac{1}{e})^{n} =\Sigma^{\infty}_{n=1} \frac{1}{e}(\frac{1}{e})^{n-1}$
$a=\frac{1}{e}$ and the common ratio is $r=\frac{1}{e}$
$|r| = |\frac{1}{e}|$
$=\frac{1}{e}$
$\approx 0.3679$
Since $|r| \lt 1$, $\Sigma^{\infty}_{n=1} (\frac{1}{e})^{n}$ is convergent and the sum is
$\Sigma^{\infty}_{n=1} (\frac{1}{e})^{n} = \frac{a}{1-r}$
$=\frac{\frac{1}{e}}{1-\frac{1}{e}}$
$=\frac{\frac{1}{e}}{\frac{(e-1)}{e}}$
$=\frac{1}{(e-1)}$
Second
$\Sigma^{\infty}_{n=1}\frac{1}{n(n+1)}$
Compute partial sums
$s_{n} =\Sigma^{\infty}_{i=1}\frac{1}{i(i+1)}$
$=\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} +...+\frac{1}{n(n+1)}$
Thus $s_{n} =\Sigma^{\infty}_{i=1}\frac{1}{i(i+1)}$
$=\Sigma^{\infty}_{i=1} (\frac{1}{i}-\frac{1}{i+1})$
Use partial fraction decomposition
$=1-\frac{1}{n+1}$
So
$\lim\limits_{n \to \infty}s_{n}=\lim\limits_{n \to \infty}(1-\frac{1}{n+1})$
$=\lim\limits_{n \to \infty}(1-\frac{\frac{1}{n}}{1+\frac{1}{n}})$
$=1-\frac{0}{1+0}$ since $\lim\limits_{n \to \infty}\frac{1}{n}=0$
$=1-0$
$=1$
Therefore the series $\Sigma^{\infty}_{n=1}\frac{1}{n(n+1)}$ is convergent and the sum is 1.
Since $\Sigma^{\infty}_{n=1} (\frac{1}{e})^{n}$ and $\Sigma^{\infty}_{n=1}\frac{1}{n(n+1)}$ are convergent, then the series is convergent and the sum is $\frac{1}{e-1}+1 = \frac{e}{e-1}$
