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Question Number 26109 by abdo imad last updated on 19/Dec/17
let s give S_n = Σ_(k=1) ^(k=n) k^(−2 ) . (k+1)^(−2) find lim_(n−>∝) S_n .
$${let}\:{s}\:{give}\:\:{S}_{{n}} \:=\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:{k}^{−\mathrm{2}\:} .\:\left({k}+\mathrm{1}\right)^{−\mathrm{2}} \\ $$$${find}\:\:{lim}_{{n}−>\propto} \:\:{S}_{{n}} \:\:. \\ $$
Commented by moxhix last updated on 19/Dec/17
((1/(k(k+1))))^2 =((1/k)−(1/(k+1)))^2 =(1/k^2 )+(1/((k+1)^2 ))−2(1/(k(k+1))) =(1/k^2 )+(1/((k+1)^2 ))−(2/k)+(2/(k+1)) S_n =Σ_(k=1) ^n ((1/k^2 )+(1/((k+1)^2 )))−2Σ_(k=1) ^n ((1/k)−(1/(k+1))) =2Σ_(k=1) ^n ((1/k^2 ))−(1/1^2 )+(1/((n+1)^2 ))−2((1/1)−(1/(n+1))) =2Σ_(k=1) ^n ((1/k^2 ))+(1/((n+1)^2 ))+(1/(n+1))−3 →2(π^2 /6)−3=(π^2 /3)−3 (n→∞) −−−−−−−−−−−−−−−−−−− show Σ_(n=1) ^∞ (1/n^2 )=(π^2 /6) (Basel probrem) ((sinx)/x)=1−(x^2 /(3!))+(x^4 /(5!))−... ((sinx)/x)=0⇒x=π,−π,2π,−2π,3π,−3π,... hence ((sinx)/x)=(1−(x/π))(1+(x/π))(1−(x/(2π)))(1+(x/(2π)))(1−(x/(3π)))(1+(x/(3π)))∙... (x^2 coefficient)=−(1/(3!))=(1/π)(((−1)/π))+(1/(2π))(((−1)/(2π)))+(1/(3π))(((−1)/(3π)))+... ∴(1/(3!))=Σ_(n=1) ^∞ (1/(n^2 π^2 )) ∴(π^2 /6)=Σ_(n=1) ^∞ (1/n^2 )
$$\left(\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)}\right)^{\mathrm{2}} =\left(\frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{{k}+\mathrm{1}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{{k}^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }−\mathrm{2}\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{{k}^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{2}}{{k}}+\frac{\mathrm{2}}{{k}+\mathrm{1}} \\ $$$$ \\ $$$${S}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}}{{k}^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }\right)−\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{{k}+\mathrm{1}}\right) \\ $$$$\:\:\:\:=\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\right)−\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right) \\ $$$$\:\:\:\:=\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\right)+\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{{n}+\mathrm{1}}−\mathrm{3} \\ $$$$\:\:\:\:\rightarrow\mathrm{2}\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\mathrm{3}=\frac{\pi^{\mathrm{2}} }{\mathrm{3}}−\mathrm{3}\:\left({n}\rightarrow\infty\right) \\ $$$$−−−−−−−−−−−−−−−−−−− \\ $$$${show}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:\:\left({Basel}\:{probrem}\right) \\ $$$$\frac{{sinx}}{{x}}=\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{5}!}−… \\ $$$$ \\ $$$$\frac{{sinx}}{{x}}=\mathrm{0}\Rightarrow{x}=\pi,−\pi,\mathrm{2}\pi,−\mathrm{2}\pi,\mathrm{3}\pi,−\mathrm{3}\pi,… \\ $$$${hence}\:\:\frac{{sinx}}{{x}}=\left(\mathrm{1}−\frac{{x}}{\pi}\right)\left(\mathrm{1}+\frac{{x}}{\pi}\right)\left(\mathrm{1}−\frac{{x}}{\mathrm{2}\pi}\right)\left(\mathrm{1}+\frac{{x}}{\mathrm{2}\pi}\right)\left(\mathrm{1}−\frac{{x}}{\mathrm{3}\pi}\right)\left(\mathrm{1}+\frac{{x}}{\mathrm{3}\pi}\right)\centerdot… \\ $$$$\left({x}^{\mathrm{2}} \:\mathrm{coefficient}\right)=−\frac{\mathrm{1}}{\mathrm{3}!}=\frac{\mathrm{1}}{\pi}\left(\frac{−\mathrm{1}}{\pi}\right)+\frac{\mathrm{1}}{\mathrm{2}\pi}\left(\frac{−\mathrm{1}}{\mathrm{2}\pi}\right)+\frac{\mathrm{1}}{\mathrm{3}\pi}\left(\frac{−\mathrm{1}}{\mathrm{3}\pi}\right)+… \\ $$$$\therefore\frac{\mathrm{1}}{\mathrm{3}!}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} \pi^{\mathrm{2}} } \\ $$$$\therefore\frac{\pi^{\mathrm{2}} }{\mathrm{6}}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$

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