Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 202123 by BaliramKumar last updated on 21/Dec/23

$$\frac{1}{1\times 3}+\frac{1}{3\times 5}+\frac{1}{5\times 7}+...............\mathrm{\infty }=?$$

Answered by Rasheed.Sindhi last updated on 21/Dec/23

$$t_n=\frac{1}{(2n-1)(2n+1)}$$$$let\frac{1}{(2n-1)(2n+1)}=\frac{a}{2n-1}+\frac{b}{2n+1}$$$$a(2n+1)+b(2n-1)=1$$$$2n(a+b)+a-b=1$$$$2(a+b)=0\wedge a-b=1\Rightarrow a=\frac{1}{2},b=-\frac{1}{2}$$$$t_n=\frac{1}{2(2n-1)}-\frac{1}{2(2n+1)}$$$$t_{n-1}=\frac{1}{2(2(n-1)-1)}-\frac{1}{2(2(n-1)+1)}$$$$t_{n-1}=\frac{1}{2(2n-2-1)}-\frac{1}{2(2n-2+1)}$$$$t_{n-1}=\frac{1}{2(2n-3)}-\frac{1}{2(2n-1)}$$$$\begin{array}{|cc|}\hline t_1& \frac{1}{2}-\cancel{\frac{1}{6}}\\ t_2& \cancel{\frac{1}{6}}-\cancel{\frac{1}{10}}\\ t_3& \cancel{\frac{1}{10}}-\cancel{\frac{1}{14}}\\ ...& ...\\ ...& ...\\ t_{n-1}& \cancel{\frac{1}{2(2n-3)}}-\cancel{\frac{1}{2(2n-1)}}\\ t_n& \cancel{\frac{1}{2(2n-1)}}-\frac{1}{2(2n+1)}\\ \mathrm{\Sigma }{t}_n& \frac{1}{2}-\frac{1}{2(2n+1)}=\frac{n}{2n+1}\\ \hline\end{array}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{t}_n=\underset{n\to \mathrm{\infty }}{\lim }\frac{n}{2n+1}=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{2+\frac{1}{n}}=\frac{1}{2}$$

Commented by BaliramKumar last updated on 21/Dec/23

$$\mathrm{Thanks}\mathrm{Sir}$$

Answered by qaz last updated on 21/Dec/23

$$\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{1}{(2k+1)(2k+3)}=\frac{1}{2}\underset{k=0}{\sum ^{\mathrm{\infty }}}(\frac{1}{2k+1}-\frac{1}{2k+3})$$$$=\frac{1}{2}\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{1}{2k+1}-\frac{1}{2}\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{2k+1}=\frac{1}{2}\sum _{k\in \left\{0\right\}}\frac{1}{2k+1}=\frac{1}{2}$$

Commented by BaliramKumar last updated on 21/Dec/23

$$\mathrm{Thanks}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com