Question Number 145082 by mathdanisur last updated on 02/Jul/21
$$\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} −\mathrm{2}}\right)=? \\ $$
Answered by phally last updated on 02/Jul/21
$$=\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\underset{\mathrm{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{n}−\mathrm{1}\right)−\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)} \\ $$$$=\left(−\frac{\mathrm{1}}{\mathrm{4}}\right)\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{n}−\mathrm{1}}\right) \\ $$$$=\left(−\frac{\mathrm{1}}{\mathrm{4}}\right)\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}−\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by mathdanisur last updated on 02/Jul/21
$${Thankyou}\:{Sir},\:{but}\:{answer}\:\frac{\mathrm{3}}{\mathrm{8}} \\ $$
Commented by phally last updated on 02/Jul/21
$$\mathrm{why}\:\mathrm{answer}\:\frac{\mathrm{3}}{\mathrm{8}}? \\ $$
Commented by mathdanisur last updated on 02/Jul/21
$${yes}\:{Ser}\:{thank}\:{you} \\ $$
Answered by Ar Brandon last updated on 02/Jul/21
![S(n)=Σ_(n=2) ^∞ (1/(2n^2 −2))=(1/2)Σ_(n≥2) (1/(n^2 −1)) =(1/2)Σ_(n≥2) (1/((n−1)(n+1)))=(1/2)Σ_(n≥2) ((1/(2(n−1)))−(1/(2(n+1)))) =(1/4)[((1/1)−(1/3))+((1/2)−(1/4))+((1/3)−(1/5))+((1/4)−(1/6))+∙∙∙] =(1/4)(1+(1/2))=(1/4)×(3/2)=(3/8)](https://www.tinkutara.com/question/Q145090.png)
$$\mathrm{S}\left(\mathrm{n}\right)=\underset{\mathrm{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2n}^{\mathrm{2}} −\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{n}\geqslant\mathrm{2}} {\sum}\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{n}\geqslant\mathrm{2}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{n}\geqslant\mathrm{2}} {\sum}\left(\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{n}−\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{n}+\mathrm{1}\right)}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\left[\left(\frac{\mathrm{1}}{\mathrm{1}}−\cancel{\frac{\mathrm{1}}{\mathrm{3}}}\right)+\left(\frac{\mathrm{1}}{\mathrm{2}}−\cancel{\frac{\mathrm{1}}{\mathrm{4}}}\right)+\left(\cancel{\frac{\mathrm{1}}{\mathrm{3}}}−\cancel{\frac{\mathrm{1}}{\mathrm{5}}}\right)+\left(\cancel{\frac{\mathrm{1}}{\mathrm{4}}}−\cancel{\frac{\mathrm{1}}{\mathrm{6}}}\right)+\centerdot\centerdot\centerdot\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{8}} \\ $$
Commented by mathdanisur last updated on 02/Jul/21
$${thank}\:{you}\:{Ser}\:{cool} \\ $$
Answered by bobhans last updated on 02/Jul/21
$$\:\:\mathrm{Find}\:\mathrm{thd}\:\mathrm{value}\:\mathrm{of}\:\underset{\mathrm{n}=\mathrm{2}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2n}^{\mathrm{2}} −\mathrm{2}}\right). \\ $$$$\underset{\mathrm{n}=\mathrm{2}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{n}^{\mathrm{2}} −\mathrm{1}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{n}=\mathrm{2}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)}\right) \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{k}\rightarrow\infty} {\mathrm{lim}}\:\underset{\mathrm{n}=\mathrm{2}} {\overset{\mathrm{k}} {\sum}}\:\left(\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{n}−\mathrm{1}}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{k}\rightarrow\infty} {\mathrm{lim}}\:\left(\cancel{\frac{\mathrm{1}}{\mathrm{3}}}−\mathrm{1}+\cancel{\frac{\mathrm{1}}{\mathrm{4}}}−\frac{\mathrm{1}}{\mathrm{2}}+\cancel{\frac{\mathrm{1}}{\mathrm{5}}}−\cancel{\frac{\mathrm{1}}{\mathrm{3}}}+\cancel{\frac{\mathrm{1}}{\mathrm{6}}}−\cancel{\frac{\mathrm{1}}{\mathrm{5}}}+…+\frac{\mathrm{1}}{\mathrm{k}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{k}−\mathrm{1}}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}.\left(−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)=\:\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{8}} \\ $$
Commented by mathdanisur last updated on 02/Jul/21
$${thankyou}\:{Ser}\:{cool} \\ $$
Answered by qaz last updated on 02/Jul/21
$$\underset{\mathrm{n}=\mathrm{2}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2n}^{\mathrm{2}} −\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{n}=\mathrm{2}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{n}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\right)=\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{3}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{x}^{\mathrm{n}} −\mathrm{x}^{\mathrm{n}+\mathrm{2}} \right)\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+\mathrm{x}\right)\mathrm{dx}=\frac{\mathrm{3}}{\mathrm{8}} \\ $$
Commented by mathdanisur last updated on 02/Jul/21
$${a}\:{lot}\:{cool}\:{Ser}\:{thank}\:{you}.. \\ $$