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Question Number 31978 by abdo imad last updated on 17/Mar/18

$$findthevalueof\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2n+1)(2n+3)(2n+5)}.$$

Answered by Joel578 last updated on 18/Mar/18

$$S_k=\underset{n=0}{\sum ^k}\frac{1}{(2n+1)(2n+3)(2n+5)}$$$$=\frac{1}{8}\underset{n=0}{\sum ^k}(\frac{1}{2n+1}-\frac{2}{2n+3}+\frac{1}{2n+5})$$$$=\frac{1}{8}[(1-\frac{2}{3}+\frac{1}{5})+(\frac{1}{3}-\frac{2}{5}+\frac{1}{7})+(\frac{1}{5}-\frac{2}{7}+\frac{1}{9})+...+(\frac{1}{2k+1}-\frac{2}{2k+3}+\frac{1}{2k+5})$$$$=\frac{1}{8}(1-\frac{1}{3}+\frac{1}{2k+5})=\frac{1}{8}(\frac{2}{3}+\frac{1}{2k+5})$$$$$$$$\underset{k\to \mathrm{\infty }}{\lim }{S}_k=\underset{k\to \mathrm{\infty }}{\lim }\frac{1}{8}(\frac{2}{3}+\frac{1}{2k+5})=\frac{1}{12}$$

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