## University Calculus: Early Transcendentals (3rd Edition)

Converges to $1$
We have $a_n=\dfrac{4}{(4n-3)(4n+1)}=\dfrac{1}{4n-3}-\dfrac{1}{4n+1}$ Here, $s_n=(1-\dfrac{1}{5})(\dfrac{1}{5}-\dfrac{1}{9})+....\dfrac{1}{4n-3}-\dfrac{1}{4n+1}=1-\dfrac{1}{4n+1}$ Thus, we have: $\lim\limits_{n \to \infty} s_n=1-\lim\limits_{n \to \infty} \dfrac{1}{4n+1}=1+0=1$ We see that $s_n \to 1$ as $n \to \infty$, so, the given series converges to $1$.