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Question Number 118488 by Lordose last updated on 18/Oct/20

$$\mathbf{Conjecture}\mathbf{a}\mathbf{formula}\mathbf{for}\mathbf{the}\mathbf{infinite}$$$$\mathbf{sum}\mathbf{of}\mathbf{the}\mathbf{series}.$$$$\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\cdot \cdot \cdot \frac{1}{(2\mathbf{n}-1)(2\mathbf{n}+1)}$$$$\mathbf{And}\mathbf{prove}\mathbf{the}\mathbf{formula}\mathbf{by}\mathbf{Induction}.$$

Answered by Olaf last updated on 18/Oct/20

$$\frac{1}{(2k-1)(2k+1)}=\frac{1}{2}(\frac{1}{2k-1}-\frac{1}{2k+1})$$$${\mathrm{S}}_n=\underset{k=1}{\sum ^n}\frac{1}{(2k-1)(2k+1)}$$$${\mathrm{S}}_n=\frac{1}{2}\underset{k=1}{\sum ^n}\frac{1}{2k-1}-\frac{1}{2}\underset{k=1}{\sum ^n}\frac{1}{2k+1}$$$${\mathrm{S}}_n=\frac{1}{2}\underset{k=0}{\sum ^{n-1}}\frac{1}{2k+1}-\frac{1}{2}\underset{k=1}{\sum ^n}\frac{1}{2k+1}$$$${\mathrm{S}}_n=\frac{1}{2}(1+\underset{k=1}{\sum ^n}\frac{1}{2k+1}-\frac{1}{2n+1})-\frac{1}{2}\underset{k=1}{\sum ^n}\frac{1}{2k+1}$$$${\mathrm{S}}_n=\frac{1}{2}(1-\frac{1}{2n+1})=\frac{n}{2n+1}$$$$$$$$\mathrm{Induction}:$$$$\mathrm{for}n=1,{\mathrm{S}}_1=\frac{1}{1.3}=\frac{1}{3}=\frac{1}{2\left(1\right)+1}$$$$\Rightarrow \mathrm{the}\mathrm{formula}\mathrm{is}\mathrm{true}\mathrm{for}n=1$$$$$$$$\mathrm{Now}\mathrm{we}\mathrm{suppose}\mathrm{the}\mathrm{formula}\mathrm{is}\mathrm{true}\mathrm{for}n.$$$${\mathrm{S}}_{n+1}={\mathrm{S}}_n+\frac{1}{(2n+1)(2n+3)}$$$${\mathrm{S}}_{n+1}=\frac{n}{2n+1}+\frac{1}{(2n+1)(2n+3)}$$$${\mathrm{S}}_{n+1}=\frac{n(2n+3)+1}{(2n+1)(2n+3)}$$$${\mathrm{S}}_{n+1}=\frac{2n^2+3n+1}{(2n+1)(2n+3)}$$$${\mathrm{S}}_{n+1}=\frac{2(n+1)(n+\frac{1}{2})}{(2n+1)(2n+3)}$$$${\mathrm{S}}_{n+1}=\frac{n+1}{2n+3}$$$${\mathrm{S}}_n\mathrm{true}\Rightarrow {\mathrm{S}}_{n+1}\mathrm{true}$$$$$$$$\mathrm{Finally},{\mathrm{S}}_n=\frac{n}{2n+1},n⩾1$$

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