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Question Number 127908 by bramlexs22 last updated on 03/Jan/21

$$\frac{1}{1.3}+\frac{2}{1.3.5}+\frac{3}{1.3.5.7}+\frac{4}{1.3.5.7.9}+...=?$$

Answered by liberty last updated on 03/Jan/21

$$\mathrm{Let}\lambda =\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{\mathrm{n}}{1.3.5...(2\mathrm{n}+1)}$$$$\mathrm{consider}\frac{\mathrm{n}}{1.3.5....(2\mathrm{n}+1)}=\frac{\mathrm{n}}{1.(3.5...(2\mathrm{n}-1)).(2\mathrm{n}+1)}$$$$=\frac{1}{2}.\frac{(2\mathrm{n}+1)-1}{1.(3.5...(2\mathrm{n}-1)).(2\mathrm{n}+1)}$$$$=\frac{1}{2}(\frac{1}{1.3.5...(2\mathrm{n}-1)}-\frac{1}{3.5....(2\mathrm{n}+1)})$$$$\mathrm{now}\mathrm{we}\mathrm{get}\lambda =\underset{\mathrm{k}\to \mathrm{\infty }}{\lim }\frac{1}{2}\underset{\mathrm{n}=1}{\sum ^{\mathrm{k}}}(\frac{1}{1.3.5....(2\mathrm{n}-1)}-\frac{1}{3.5....(2\mathrm{n}+1)})$$$$\mathrm{telescoping}\mathrm{series}$$$$\lambda =\underset{\mathrm{k}\to \mathrm{\infty }}{\lim }\frac{1}{2}(\frac{1}{1}-\frac{1}{3.5....(2\mathrm{k}+1)})$$$$\lambda =\frac{1}{2}(1-0)=\frac{1}{2}$$$$$$

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