find-the-value-of-n-0-1-2n-1-2n-3-

Question Number 31977 by abdo imad last updated on 17/Mar/18

f i n d t h e v a l u e o f \sum_{n = 0}^{\infty} \frac{1}{(2 n + 1) (2 n + 3)}

Commented by Tinkutara last updated on 17/Mar/18

\frac{1}{2}

Commented by abdo imad last updated on 18/Mar/18

l e t p u t S_{n} = \sum_{k = 0}^{n} \frac{1}{(2 k + 1) (2 k + 3)} w e h a v e

S_{n} = \frac{1}{2} \sum_{k = 0}^{n} (\frac{1}{2 k + 1} - \frac{1}{2 k + 3}) \Rightarrow

2 S_{n} = \sum_{k = 0}^{n} (a_{k} - a_{k + 1}) w i t h a_{k} = \frac{1}{2 k + 1}

= a_{0} - a_{1} + a_{1} - a_{2} + \dots . + a_{n} - a_{n + 1} = a_{0} - a_{n + 1} = 1 - \frac{1}{2 n + 3}

\Rightarrow S_{n} = \frac{1}{2} - \frac{1}{2 (2 n + 3)} \Rightarrow {l i m}_{n \to \infty} S_{n} = \frac{1}{2} .

Answered by Joel578 last updated on 18/Mar/18

S_{k} = \underset{n = 0}{\sum^{k}} \frac{1}{(2 n + 1) (2 n + 3)}

= \frac{1}{2} \underset{n = 0}{\sum^{k}} (\frac{1}{2 n + 1} - \frac{1}{2 n + 3})

= \frac{1}{2} [(1 - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{5}) + \dots + (\frac{1}{2 k + 1} - \frac{1}{2 k + 3})

= \frac{1}{2} (1 - \frac{1}{2 k + 3})

\lim_{k \to \infty} S_{k} = \lim_{k \to \infty} \frac{1}{2} (1 - \frac{1}{2 k + 3}) = \frac{1}{2}

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