Answer
converges to $1$.
Work Step by Step
consider $a_n=\dfrac{4}{(4n-3)(4n+1)}=\dfrac{1}{4n-3}-\dfrac{1}{4n+1}$
The nth partial sums are: $ 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})$
Now, apply limits $\lim\limits_{n \to \infty} s_n=(1)-\lim\limits_{n \to \infty} \dfrac{1}{4n+1}=(1)+(0)=1$
Thus,
As $n \to \infty$ then $s_n \to 1$ .
Hence, the series converges to $1$.
