Question

$\text{Given}{S}_1=\sum _{n=0}^k\frac{2n+3}{(n+1{)}^2(n+2{)}^2}\text{and}{S}_2=\sum _{n=1}^k\frac{2}{4n^2-1}$

$\text{Evaluate}\underset{k\to \infty }{lim}{S}_1-S_2$

• A. $0$
• B. $\frac{1}{12}$
• C. $\frac{1}{3}$
• D. $\frac{1}{4}$

Answer 1

The correct option is A $0$

$S_1=\sum _{n=0}^k\frac{2n+3}{(n+1{)}^2(n+2{)}^2}=\sum _{n=0}^k\frac{1}{(n+1{)}^2}-\frac{1}{(n+2{)}^2}=(\frac{1}{1}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{9})+(\frac{1}{9}-\frac{1}{16})+...+(\frac{1}{(k+1{)}^2}-\frac{1}{(k+2{)}^2})=1-\frac{1}{(k+2{)}^2}\text{Now,}k\to \infty S_1=\underset{k\to \infty }{lim}(1-\frac{1}{(k+2{)}^2})=1$

$S_2=\sum _{n=1}^k\frac{2}{4n^2-1}=\sum _{n=1}^k\frac{1}{2n-1}-\frac{1}{2n+1}=(\frac{1}{1}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{7})+...+(\frac{1}{2k-1}-\frac{1}{2k+1})=1-\frac{1}{2k+1}\text{Now,}k\to \infty S_2=\underset{k\to \infty }{lim}(1-\frac{1}{2k+1})=1\therefore S_1-S_2=1-1=0$