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Question Number 8764 by tawakalitu last updated on 26/Oct/16

$$\left(\mathrm{a}\right)\mathrm{Find}\mathrm{the}\mathrm{sum}\mathrm{given}\mathrm{by}$$$${\mathrm{S}}_{\mathrm{n}}=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{(2\mathrm{n}-1)(2\mathrm{n}+1)}$$$$\left(\mathrm{b}\right)\mathrm{find}\mathrm{the}\mathrm{limit}\mathrm{of}{\mathrm{S}}_{\mathrm{n}}\mathrm{as}\mathrm{n}\to \mathrm{\infty }$$

Commented by sou1618 last updated on 26/Oct/16

$$S_n=\frac{1}{2}\{(\frac{1}{1}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{5})+....+(\frac{1}{2n-1}-\frac{1}{2n+1})\}$$$$=\frac{1}{2}(1-\frac{1}{2n+1})$$$$=\frac{n}{2n+1}$$$${lim}_{n\to \mathrm{\infty }}\frac{n}{2n+1}={lim}_{n\to \mathrm{\infty }}\frac{1}{(2+\frac{1}{n})}=\frac{1}{2}$$

Commented by tawakalitu last updated on 26/Oct/16

$$\mathrm{Thank}\mathrm{you}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}.$$

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