Answer
$\Sigma^{\infty}_{n=1}a_{n}=\frac{1}{4}$
Work Step by Step
$s_{n} = \frac{n^{2}-1}{4n^{2}+1}$
$\lim\limits_{n \to \infty}s_{n} = \lim\limits_{n \to \infty} (\frac{n^{2}-1}{4n^{2}+1})$
$\lim\limits_{n \to \infty}(\frac{1-\frac{1}{n^{2}}}{4+\frac{1}{n^{2}}})$
$=\frac{1}{4}$
$\Sigma^{\infty}_{n=1}a_{n}=\frac{1}{4}$
