Answer
$\dfrac{-2}{9}$
Work Step by Step
Consider $s_n=\dfrac{-8}{(4n-3)(4n+1)}$
Re-write the given series as: $s_n=\dfrac{-8}{(4n-3)(4n+1)}=\dfrac{-2}{(4n-3))}+\dfrac{2}{(4n+1)}$
$s_n=(\dfrac{-2}{9}+\dfrac{2}{13})+(\dfrac{-2}{13}+\dfrac{2}{17})+(\dfrac{-2}{17}+\dfrac{2}{21})+(\dfrac{-2}{21}+\dfrac{-2}{25})...+(\dfrac{-2}{4n-3}+\dfrac{2}{4n+1})$
or,
$s_n=\dfrac{-2}{9}+\dfrac{2}{4n+1}$
Apply limits, we get:
$\lim\limits_{n \to \infty}s_n=\lim\limits_{n \to \infty}(\dfrac{-2}{9}+\dfrac{2}{4n+1})$
After simplifications, we get
$\lim\limits_{n \to \infty}s_n=(\dfrac{-2}{9}+0)$
Hence, $\lim\limits_{n \to \infty}s_n=\dfrac{-2}{9}$
