Question Number 137403 by BHOOPENDRA last updated on 02/Apr/21
Commented by BHOOPENDRA last updated on 02/Apr/21
$${please}\:{help}\:{me}\:{out}\:{this}\:{in}\:{correct}\:{way} \\ $$
Answered by liberty last updated on 02/Apr/21
![(1/(1.3)) +(1/(3.5))+(1/(5.7))+...= (1/3)+Σ_(n=1) ^∞ (1/((2n+1)(2n+3))) =(1/3)+lim_(n→∞) ((1/(3.5))+(1/(5.7))+(1/(7.9))+...+(1/((2n+1)(2n+3))) =(1/3)+lim_(n→∞) (1/2)[ ((5−3)/(3.5))+((7−5)/(5.7))+((9−7)/(7.9))+...+(((2n+3)−(2n+1))/((2n+1)(2n+3))) ] = (1/3)+lim_(n→∞) (1/2) [ (1/3)−(1/5)+(1/5)−(1/7)+(1/7)−(1/9)+...+(1/(2n+1))−(1/(2n+3)) ] =(1/3)+(1/2) lim_(n→∞) [ (1/3)−(1/(2n+3)) ] = (1/3)+(1/6)= (1/2)](https://www.tinkutara.com/question/Q137414.png)
$$\frac{\mathrm{1}}{\mathrm{1}.\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{3}.\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{5}.\mathrm{7}}+…=\:\frac{\mathrm{1}}{\mathrm{3}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}+\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{3}.\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{5}.\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{7}.\mathrm{9}}+…+\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)}\right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}+\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{2}}\left[\:\frac{\mathrm{5}−\mathrm{3}}{\mathrm{3}.\mathrm{5}}+\frac{\mathrm{7}−\mathrm{5}}{\mathrm{5}.\mathrm{7}}+\frac{\mathrm{9}−\mathrm{7}}{\mathrm{7}.\mathrm{9}}+…+\frac{\left(\mathrm{2}{n}+\mathrm{3}\right)−\left(\mathrm{2}{n}+\mathrm{1}\right)}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)}\:\right] \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{3}}+\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{2}}\:\left[\:\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{7}}−\frac{\mathrm{1}}{\mathrm{9}}+…+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{3}}\:\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left[\:\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{3}}\:\right] \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by BHOOPENDRA last updated on 02/Apr/21
$${what}\:{about}\:{the}\:{upper}\:{part}\:{of}\:{the}\:{question}\:{sir} \\ $$
Commented by BHOOPENDRA last updated on 02/Apr/21
$${you}\:{have}\:{to}\:{express}\:{in}\:{I}\:{form}\: \\ $$
Commented by BHOOPENDRA last updated on 02/Apr/21
$${read}\:{the}\:{question}\:{again}\:,{you}\:{are}\:{devlope} \\ $$$${f}\left({x}\right){as}\:{afourier}\:{series}\:{and}\:{then}\: \\ $$$${evaluate}\:{this}\:{f}\left({x}\right)={i}\left({x}\right){here} \\ $$