Question

${lim}_{n\to \infty }(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+....n\text{terms})=$

• A. $\frac{1}{3}$
• B. $\frac{1}{12}$
• C. $\frac{1}{7}$
• D. $\frac{1}{10}$

Answer 1

Answer: (B)

$\frac{1}{12}$

Above series can be rewritten as

${\sum }_{n=1}^{\infty }\frac{1}{(4n-1)\times (4n+3)}={\sum }_{n=1}^{\infty }\frac{(4n+3)-(4n-1)}{4\times (4n-1)(4n+3)}=\frac{1}{4}{\sum }_{n=1}^{\infty }(\frac{1}{4n-1}-\frac{1}{4n+3})={lim}_{n\to \infty }\frac{1}{4}\times (\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-...........-\frac{1}{4n-1}+\frac{1}{4n-1}-\frac{1}{4n+3})={lim}_{n\to \infty }\frac{1}{4}\times (\frac{1}{3}-\frac{1}{4n+3})=\frac{1}{4}\times (\frac{1}{3}-0)=\frac{1}{12}$