Question

$\underset{n\to \infty }{\lim }\left\{\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2n+1\right)\left(2n+3\right)}\right\}$ is equal to

(a) 0
(b) 1/2
(c) 1/9
(d) 2

Answer 1

(b) 1/2

$$\begin{gathered} \underset{n\to \infty }{\lim }\left[\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}...\frac{1}{\left(2n+1\right)\left(2n+3\right)}\right] \\ \mathrm{Here},{\mathrm{T}}_n=\frac{1}{\left(2n-1\right)\left(2n+1\right)} \\ \Rightarrow {\mathrm{T}}_n=\frac{\mathrm{A}}{\left(2n-1\right)}+\frac{\mathrm{B}}{\left(2n+1\right)} \\ \mathrm{On}\mathrm{equating}\mathrm{A}=\frac{1}{2}\mathrm{and}\mathrm{B}=-\frac{1}{2}: \\ {\mathrm{T}}_n=\frac{1}{2\left(2n-1\right)}-\frac{1}{2\left(2n+1\right)} \\ \Rightarrow {\mathrm{T}}_1=\frac{1}{2}\left[1-\frac{1}{3}\right] \\ \Rightarrow {\mathrm{T}}_2=\frac{1}{2}\left[\frac{1}{3}-\frac{1}{5}\right] \\ \Rightarrow {\mathrm{T}}_{n-1}=\frac{1}{2}\left[\frac{1}{2n-1}-\frac{1}{2n-1}\right] \\ \Rightarrow {\mathrm{T}}_n=\frac{1}{2}\left[\frac{1}{2n-1}-\frac{1}{2n+1}\right] \\ \Rightarrow {\mathrm{T}}_1+{\mathrm{T}}_2+{\mathrm{T}}_3...{\mathrm{T}}_n=\frac{1}{2}\left[1-\frac{1}{2n+1}\right] \\ \Rightarrow {\mathrm{T}}_1+{\mathrm{T}}_2+{\mathrm{T}}_3...{\mathrm{T}}_n=\frac{1}{2}\left[\frac{2n}{2n+1}\right] \\ \Rightarrow {\mathrm{T}}_1+{\mathrm{T}}_2+{\mathrm{T}}_3...{\mathrm{T}}_n=\frac{n}{2n+1} \\ \therefore \underset{n\to \infty }{\lim }\left[\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}...\frac{1}{\left(2n+1\right)\left(2n+3\right)}\right] \\ =\underset{n\to \infty }{\lim }\left[\sum _{n=1}^n\frac{1}{\left(2n-1\right)\left(2n+1\right)}\right] \\ =\underset{n\to \infty }{\lim }\left(\frac{n}{2n+1}\right) \end{gathered}$$

$$\begin{gathered} =\underset{n\to \infty }{\lim }\left(\frac{1}{2{\displaystyle }+{\displaystyle \frac{1}{n}}}\right)\left[\mathrm{Divid}\text{ing}{\mathrm{N}}^r\text{and}{\mathrm{D}}^r\mathrm{by}n\right] \\ =\frac{1}{2} \end{gathered}$$