## Thomas' Calculus 13th Edition

$\dfrac{-2}{9}$
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, $s_n=\dfrac{-2}{9}+\dfrac{2}{4n+1}$ Applying 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, $\lim\limits_{n \to \infty}s_n=\dfrac{-2}{9}$