#### Answer

a. $\left\{\frac{2}{3},\frac{4}{5},\frac{6}{7},\frac{8}{9}\right\}$
b.$S_n=\frac{2n}{2n+1}$
c. 1

#### Work Step by Step

a. $\frac{2}{(2(1)-1)(2(1)+1)}=\frac{2}{3}$
$\frac{2}{(2(2)-1)(2(2)+1)}=\frac{2}{15}$
$\frac{2}{15}+\frac{2}{3}=\frac{12}{15}=\frac{4}{5}$
$\frac{2}{(2(3)-1)(2(3)+1)}=\frac{2}{35}$
$\frac{2}{35}+\frac{4}{5}=\frac{30}{35}=\frac{6}{7}$
$\frac{2}{(2(4)-1)(2(4)+1)}=\frac{2}{63}$
$\frac{2}{63}+\frac{6}{7}=\frac{56}{63}=\frac{8}{9}$
b. By looking at the $S_n$ values, one notices each time $n$ increases by one, both the numerator and denominator increase by 2. The numerator is always $2x$ the $n$ value, and the denominator is one greater than the numerator. Therefore, $S_n=\frac{2n}{2n+1}$
3. The value of the series is $\lim\limits_{n \to \infty}S_n$, which is $\lim\limits_{n \to \infty}\frac{2n}{2n+1}$, which equals 1 because the numerator and denominator have the same order and their leading coefficients are the same.