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