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Question Number 202123 by BaliramKumar last updated on 21/Dec/23
(1/(1×3)) + (1/(3×5)) + (1/(5×7)) + ...............∞ = ?
$$\frac{\mathrm{1}}{\mathrm{1}×\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{3}×\mathrm{5}}\:+\:\frac{\mathrm{1}}{\mathrm{5}×\mathrm{7}}\:+\:……………\infty\:=\:? \\ $$
Answered by Rasheed.Sindhi last updated on 21/Dec/23
t_n =(1/((2n−1)(2n+1))) let (1/((2n−1)(2n+1)))=(a/(2n−1))+(b/(2n+1)) a(2n+1)+b(2n−1)=1 2n(a+b)+a−b=1 2(a+b)=0 ∧ a−b=1⇒a=(1/2) ,b=−(1/2) t_n =(1/(2(2n−1)))−(1/(2(2n+1))) t_(n−1) =(1/(2(2 (n−1)−1)))−(1/(2(2(n−1)+1))) t_(n−1) =(1/(2(2n−2−1)))−(1/(2(2n−2+1))) t_(n−1) =(1/(2(2n−3)))−(1/(2(2n−1))) determinant ((t_1 ,((1/2)−(1/6))),(t_2 ,((1/6)−(1/(10)))),(t_3 ,((1/(10))−(1/(14)))),((...),(...)),((...),(...)),(t_(n−1) ,((1/(2(2n−3)))−(1/(2(2n−1))))),(t_n ,((1/(2(2n−1)))−(1/(2(2n+1))))),((Σt_n ),((1/2)−(1/(2(2n+1)))=(n/(2n+1))))) Σ_(n=1) ^∞ t_n =lim_(n→∞) (n/(2n+1))=lim_(n→∞) (1/(2+(1/n)))=(1/2)
$${t}_{{n}} =\frac{\mathrm{1}}{\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)} \\ $$$$\:\:\:\:\:{let}\:\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}=\frac{{a}}{\mathrm{2}{n}−\mathrm{1}}+\frac{{b}}{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\:\:\:{a}\left(\mathrm{2}{n}+\mathrm{1}\right)+{b}\left(\mathrm{2}{n}−\mathrm{1}\right)=\mathrm{1} \\ $$$$\:\:\:\mathrm{2}{n}\left({a}+{b}\right)+{a}−{b}=\mathrm{1} \\ $$$$\:\:\:\:\mathrm{2}\left({a}+{b}\right)=\mathrm{0}\:\wedge\:{a}−{b}=\mathrm{1}\Rightarrow{a}=\frac{\mathrm{1}}{\mathrm{2}}\:,{b}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:{t}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{n}−\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right)} \\ $$$$\:\:\:{t}_{{n}−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}\:\left({n}−\mathrm{1}\right)−\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}\left({n}−\mathrm{1}\right)+\mathrm{1}\right)} \\ $$$$\:\:\:{t}_{{n}−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{n}−\mathrm{2}−\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{n}−\mathrm{2}+\mathrm{1}\right)} \\ $$$$\:\:\:{t}_{{n}−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{n}−\mathrm{3}\right)}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{n}−\mathrm{1}\right)} \\ $$$$\begin{array}{|c|c|c|c|c|c|c|c|}{{t}_{\mathrm{1}} }&\hline{\frac{\mathrm{1}}{\mathrm{2}}−\cancel{\frac{\mathrm{1}}{\mathrm{6}}}}\\{{t}_{\mathrm{2}} }&\hline{\cancel{\frac{\mathrm{1}}{\mathrm{6}}}−\cancel{\frac{\mathrm{1}}{\mathrm{10}}}}\\{{t}_{\mathrm{3}} }&\hline{\cancel{\frac{\mathrm{1}}{\mathrm{10}}}−\cancel{\frac{\mathrm{1}}{\mathrm{14}}}}\\{…}&\hline{…}\\{…}&\hline{…}\\{{t}_{{n}−\mathrm{1}} }&\hline{\cancel{\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{n}−\mathrm{3}\right)}}−\cancel{\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{n}−\mathrm{1}\right)}}}\\{{t}_{{n}} }&\hline{\cancel{\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{n}−\mathrm{1}\right)}}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right)}}\\{\Sigma{t}_{{n}} }&\hline{\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right)}=\frac{{n}}{\mathrm{2}{n}+\mathrm{1}}}\\\hline\end{array} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{t}_{{n}} =\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}}{\mathrm{2}{n}+\mathrm{1}}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{2}+\frac{\mathrm{1}}{{n}}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by BaliramKumar last updated on 21/Dec/23
Thanks Sir
$$\mathrm{Thanks}\:\mathrm{Sir} \\ $$
Answered by qaz last updated on 21/Dec/23
Σ_(k=0) ^∞ (1/((2k+1)(2k+3)))=(1/2)Σ_(k=0) ^∞ ((1/(2k+1))−(1/(2k+3))) =(1/2)Σ_(k=0) ^∞ (1/(2k+1))−(1/2)Σ_(k=1) ^∞ (1/(2k+1))=(1/2)Σ_(k∈{0}) (1/(2k+1))=(1/2)
$$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{3}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{3}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}\in\left\{\mathrm{0}\right\}} {\sum}\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by BaliramKumar last updated on 21/Dec/23
Thanks
$$\mathrm{Thanks} \\ $$

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