## Calculus 8th Edition

$a_{1} = 0$ and for $n \gt 1$, $a_{n} = \frac{2}{n^{2}+n}$ $\Sigma^{\infty}_{n=1} a_{n}=1$
$s_{n} = \frac{n-1}{n+1}$ If $n=1, a_{1}=s_{1} = \frac{1-1}{1+1} = 0$ For $n \gt 1$ (or $n \geq 2$), $a_{n} = s_{n} - s_{n-1}$ $=\frac{n-1}{n+1}-\frac{(n-1)-1}{(n-1)+1}$ $=\frac{n-1}{n+1}-\frac{n-2}{n}$ $=\frac{n(n-1)-(n+1)(n-2)}{n(n+1)}$ $=\frac{n(n-1)-(n+1)(n-2)}{n^{2} + n}$ $=\frac{2}{n^{2}+n}$ Now, $\Sigma^{\infty}_{n=1}a_{n} = \lim\limits_{n \to \infty}s_{n}$ $= \lim\limits_{n \to \infty} \frac{n-1}{n+1}$ $= \lim\limits_{n \to \infty} \frac{n}{n}$ $=1$ Hence, $a_{1} = 0$ and for $n \gt 1$, $a_{n} = \frac{2}{n^{2}+n}$ $\Sigma^{\infty}_{n=1} a_{n}=1$