Question

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

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

Answer 1

Answer: (A)

Given,
$\underset{n\to \infty }{\lim }\left(\frac{1}{3\cdot 7}+\frac{1}{7\cdot 11}+\frac{1}{11\cdot 15}+\dots +(n\text{ terms })\right)$
$=\underset{n\to \infty }{\lim }\frac{1}{4}\left(\frac{4}{3\cdot 7}+\frac{4}{7\cdot 11}+\frac{4}{11\cdot 15}+\dots \right)$
$$\begin{gathered} =\underset{n\to \infty }{\lim }\frac{1}{4}(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}\dots \\ -\frac{1}{n}+\frac{1}{n}-\frac{1}{n+4}) \end{gathered}$$
$=\underset{n\to \infty }{\lim }\frac{1}{4}\left(\frac{1}{3}-\frac{1}{n+4}\right)$
$=\underset{n\to \infty }{\lim }\frac{1}{12}-\underset{n\to \infty }{\lim }\frac{1}{4(n+4)}$
$=\frac{1}{12}-\underset{n\to \infty }{\lim }\frac{1}{4n\left(1+\frac{4}{n}\right)}=\frac{1}{12}-0=\frac{1}{12}$