## Calculus 8th Edition

The series is convergent and the sum is $\frac{e}{e-1}$.
$\Sigma^{\infty}_{n=1} (\frac{1}{e^{n}}+\frac{1}{n(n+1)})$ $=\Sigma^{\infty}_{n=1} \frac{1}{e^{n}} + \Sigma^{\infty}_{n=1}\frac{1}{n(n+1)}$ $=\Sigma^{\infty}_{n=1} (\frac{1}{e})^{n} + \Sigma^{\infty}_{n=1}\frac{1}{n(n+1)}$ 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 $\frac{1}{e-1}+1 = \frac{e}{e-1}$