Question

Solve the given limit:

$\displaystyle\lim_{n \to \infty} \sum_{k=0}^n \dfrac{1}{(2k+1)(2k+3)}$

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \displaystyle\lim_{n \to \infty} \sum_{k=0}^n \dfrac{1}{(2k+1)(2k+3)}$

can be rewritten as

$\rm \: = \: \dfrac{1}{2} \displaystyle\lim_{n \to \infty} \sum_{k=0}^n \dfrac{2}{(2k+1)(2k+3)}$

$\rm \: = \: \dfrac{1}{2} \displaystyle\lim_{n \to \infty} \sum_{k=0}^n \dfrac{3 - 1}{(2k+1)(2k+3)}$

$\rm \: = \: \dfrac{1}{2} \displaystyle\lim_{n \to \infty} \sum_{k=0}^n \dfrac{2k + 3 - 2k - 1}{(2k+1)(2k+3)}$

$\rm \: = \: \dfrac{1}{2} \displaystyle\lim_{n \to \infty} \sum_{k=0}^n \dfrac{(2k + 3) - (2k + 1)}{(2k+1)(2k+3)}$

$\rm \: = \: \dfrac{1}{2} \displaystyle\lim_{n \to \infty} \sum_{k=0}^n \bigg(\dfrac{1}{2k + 1} - \dfrac{1}{2k + 3} \bigg)$

$\rm \: = \: \dfrac{1}{2} \displaystyle\lim_{n \to \infty} \bigg[ \bigg(\dfrac{1}{1} - \dfrac{1}{3} \bigg) + \bigg(\dfrac{1}{3} - \dfrac{1}{5} \bigg) + \bigg(\dfrac{1}{5} - \dfrac{1}{7} \bigg) + - - - + \bigg(\dfrac{1}{2n + 1} - \dfrac{1}{2n + 3} \bigg)$

$\rm \: = \: \dfrac{1}{2} \displaystyle\lim_{n \to \infty} \bigg[ \bigg(1 - \dfrac{1}{2n + 3} \bigg)$

$\rm \: = \: \dfrac{1}{2} \times (1 - 0)$

$\rm \: = \: \dfrac{1}{2}$

Hence,

$\rm\implies \:\boxed{\sf{  \:\rm \: \displaystyle\lim_{n \to \infty} \sum_{k=0}^n \dfrac{1}{(2k+1)(2k+3)} = \frac{1}{2} \: \: }}$

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Additional Information :-

$\boxed{\sf{  \:\displaystyle\lim_{x \to 0} \rm \: \frac{sinx}{x} \: = \: 1 \: }}$

$\boxed{\sf{  \:\displaystyle\lim_{x \to 0} \rm \: \frac{tanx}{x} \: = \: 1 \: }}$

$\boxed{\sf{  \:\displaystyle\lim_{x \to 0} \rm \: \frac{log(1 + x)}{x} \: = \: 1 \: }}$

$\boxed{\sf{  \:\displaystyle\lim_{x \to 0} \rm \: \frac{ {e}^{x} - 1}{x} \: = \: 1 \: }}$

$\boxed{\sf{  \:\displaystyle\lim_{x \to 0} \rm \: \frac{ {a}^{x} - 1}{x} \: = \: loga \: }}$