Question

\[\lim_{n \to \infty} \left\{ \frac{1}{1 . 3} + \frac{1}{3 . 5} + \frac{1}{5 . 7} + . . . + \frac{1}{\left( 2n + 1 \right) \left( 2n + 3 \right)} \right\}\]is equal to

विकल्प

•  0 

•  1/2 

• 1/9

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Answer 1

1/2 

\[\lim_{n \to \infty} \left[ \frac{1}{1 . 3} + \frac{1}{3 . 5} + \frac{1}{5 . 7} . . . \frac{1}{\left( 2n + 1 \right) \left( 2n + 3 \right)} \right]\]
\[\text{ Here }, T_n = \frac{1}{\left( 2n - 1 \right) \left( 2n + 1 \right)}\]
\[ \Rightarrow T_n = \frac{A}{\left( 2n - 1 \right)} + \frac{B}{\left( 2n + 1 \right)}\]
\[\text{ On equating } A = \frac{1}{2} \text{ and } B = - \frac{1}{2}: \]
\[ T_n = \frac{1}{2\left( 2n - 1 \right)} - \frac{1}{2\left( 2n + 1 \right)}\]
\[ \Rightarrow T_1 = \frac{1}{2}\left[ 1 - \frac{1}{3} \right]\]
\[ \Rightarrow T_2 = \frac{1}{2}\left[ \frac{1}{3} - \frac{1}{5} \right]\]
\[ \Rightarrow T_{n - 1} = \frac{1}{2}\left[ \frac{1}{2n - 1} - \frac{1}{2n - 1} \right]\]
\[ \Rightarrow T_n = \frac{1}{2}\left[ \frac{1}{2n - 1} - \frac{1}{2n + 1} \right]\]
\[ \Rightarrow T_1 + T_2 + T_3 . . . T_n = \frac{1}{2}\left[ 1 - \frac{1}{2n + 1} \right]\]
\[ \Rightarrow T_1 + T_2 + T_3 . . . T_n = \frac{1}{2}\left[ \frac{2n}{2n + 1} \right]\]
\[ \Rightarrow T_1 + T_2 + T_3 . . . T_n = \frac{n}{2n + 1}\]
\[ \therefore \lim_{n \to \infty} \left[ \frac{1}{1 . 3} + \frac{1}{3 . 5} + \frac{1}{5 . 7} . . . \frac{1}{\left( 2n + 1 \right) \left( 2n + 3 \right)} \right]\]
\[ = \lim_{n \to \infty} \left[ \sum^n_{n = 1} \frac{1}{\left( 2n - 1 \right) \left( 2n + 1 \right)} \right]\]
\[ = \lim_{n \to \infty} \left( \frac{n}{2n + 1} \right)\]

\[ = \lim_{n \to \infty} \left( \frac{1}{2 + \frac{1}{n}} \right) \left[ \text{ Dividing } N^r and D^r \text{ by } n \right]\]
\[ = \frac{1}{2}\]