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Question-120937

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Question Number 120937 by Khalmohmmad last updated on 04/Nov/20
Answered by Dwaipayan Shikari last updated on 04/Nov/20
Σ_(n=1) ^∞ ((2n+1)/((n+1)^2 (n+2)^2 )) Σ_(n=1) ^∞ (1/((n+1)^2 ))−(1/((n+2)^2 ))−(2/(((n+1)(n+2))^2 )) =(π^2 /6)−1−(π^2 /6)+(1+(1/4))−2Σ_(n=1) ^∞ ((1/((n+1)^2 ))+(1/((n+2)^2 ))−(2/((n+1)(n+2)))) =(1/4)−2Σ_(n=1) ^∞ ((π^2 /6)−1+(π^2 /6)−1−(1/4)−)+4Σ_(n=1) ^∞ (1/((n+1)))−(1/((n+2))) =(1/4)−((2π^2 )/3)+(9/2)+4((1/2)−0) =((27)/4)−((2π^2 )/3)
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}{n}+\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} \left({n}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)^{\mathrm{2}} }−\frac{\mathrm{2}}{\left(\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\right)^{\mathrm{2}} } \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\mathrm{1}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}+\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\right)−\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)^{\mathrm{2}} }−\frac{\mathrm{2}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\mathrm{1}+\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}−\right)+\mathrm{4}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{3}}+\frac{\mathrm{9}}{\mathrm{2}}+\mathrm{4}\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{0}\right) \\ $$$$=\frac{\mathrm{27}}{\mathrm{4}}−\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{3}} \\ $$
Answered by mathmax by abdo last updated on 05/Nov/20
S_n =Σ_(k=1) ^n ((2k+1)/((k+1)^2 (k+2)^2 )) ⇒ S_n =_(k+1=p) Σ_(p=2) ^(n+1) ((2(p−1)+1)/(p^2 (p+1)^2 )) =Σ_(p=2) ^(n+1) ((2p−1)/((p(p+1))^2 )) let decompose F(x) =((2x−1)/(x^2 (x+1)^2 )) ⇒F(x)=(a/x)+(b/x^2 )+(c/(x+1))+(d/((x+1)^2 )) b=−1 , d=−3 ⇒F(x)=(a/x)−(1/x^2 ) +(c/(x+1)) −(3/((x+1)^2 )) lim_(x→+∞) xF(x)=0 =a+c ⇒c=−a ⇒F(x)=(a/x)−(1/x^2 )−(a/(x+1))−(3/((x+1)^2 )) F(1)=(1/4) =a−1−(a/2)−(3/4)=(a/2)−(7/4) ⇒(a/2)=2 ⇒a=4 ⇒ F(x)=(4/x)−(4/(x+1))−(1/x^2 )−(3/((x+1)^2 )) ⇒ S_n =Σ_(k=2) ^(n+1) ((4/k)−(4/(k+1))−(1/k^2 )−(3/((k+1)^2 ))) =4 Σ_(k=2) ^(n+1) ((1/k)−(1/(k+1))) −Σ_(k=2) ^(n+1) (1/k^2 )−3 Σ_(k=2) ^(n+1) (1/((k+1)^2 )) =4((1/2)−(1/3)+(1/3)−(1/4)+...+(1/(n+1))−(1/(n+2)))−Σ_(k=1) ^(n+1) (1/k^2 ) +1−3Σ_(k=3) ^(n+2) (1/k^2 ) =4((1/2)−(1/(n+2)))−ξ_(n+1) (2)+1−3(ξ_(n+2) (2)−1−(1/4)) ⇒ lim_(n→+∞) S_n =2−(π^2 /6)+1−3((π^2 /6)−(5/4)) =3−(π^2 /6)−((3π^2 )/6) +((15)/4) =((27)/4)−((2π^2 )/3)
$$\mathrm{S}_{\mathrm{n}} =\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{2k}+\mathrm{1}}{\left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{k}+\mathrm{2}\right)^{\mathrm{2}} }\:\Rightarrow\:\mathrm{S}_{\mathrm{n}} =_{\mathrm{k}+\mathrm{1}=\mathrm{p}} \:\sum_{\mathrm{p}=\mathrm{2}} ^{\mathrm{n}+\mathrm{1}} \:\frac{\mathrm{2}\left(\mathrm{p}−\mathrm{1}\right)+\mathrm{1}}{\mathrm{p}^{\mathrm{2}} \left(\mathrm{p}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\sum_{\mathrm{p}=\mathrm{2}} ^{\mathrm{n}+\mathrm{1}} \:\frac{\mathrm{2p}−\mathrm{1}}{\left(\mathrm{p}\left(\mathrm{p}+\mathrm{1}\right)\right)^{\mathrm{2}} }\:\:\mathrm{let}\:\mathrm{decompose}\:\mathrm{F}\left(\mathrm{x}\right)\:=\frac{\mathrm{2x}−\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{\mathrm{x}}+\frac{\mathrm{b}}{\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{c}}{\mathrm{x}+\mathrm{1}}+\frac{\mathrm{d}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{b}=−\mathrm{1}\:\:,\:\mathrm{d}=−\mathrm{3}\:\Rightarrow\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{\mathrm{x}}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\:+\frac{\mathrm{c}}{\mathrm{x}+\mathrm{1}}\:−\frac{\mathrm{3}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \mathrm{xF}\left(\mathrm{x}\right)=\mathrm{0}\:=\mathrm{a}+\mathrm{c}\:\Rightarrow\mathrm{c}=−\mathrm{a}\:\Rightarrow\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{\mathrm{x}}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }−\frac{\mathrm{a}}{\mathrm{x}+\mathrm{1}}−\frac{\mathrm{3}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{F}\left(\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{4}}\:=\mathrm{a}−\mathrm{1}−\frac{\mathrm{a}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{4}}=\frac{\mathrm{a}}{\mathrm{2}}−\frac{\mathrm{7}}{\mathrm{4}}\:\Rightarrow\frac{\mathrm{a}}{\mathrm{2}}=\mathrm{2}\:\Rightarrow\mathrm{a}=\mathrm{4}\:\Rightarrow \\ $$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{4}}{\mathrm{x}}−\frac{\mathrm{4}}{\mathrm{x}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }−\frac{\mathrm{3}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\mathrm{S}_{\mathrm{n}} =\sum_{\mathrm{k}=\mathrm{2}} ^{\mathrm{n}+\mathrm{1}} \:\left(\frac{\mathrm{4}}{\mathrm{k}}−\frac{\mathrm{4}}{\mathrm{k}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }−\frac{\mathrm{3}}{\left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{2}} }\right) \\ $$$$=\mathrm{4}\:\sum_{\mathrm{k}=\mathrm{2}} ^{\mathrm{n}+\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{k}}−\frac{\mathrm{1}}{\mathrm{k}+\mathrm{1}}\right)\:−\sum_{\mathrm{k}=\mathrm{2}} ^{\mathrm{n}+\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }−\mathrm{3}\:\sum_{\mathrm{k}=\mathrm{2}} ^{\mathrm{n}+\mathrm{1}} \:\frac{\mathrm{1}}{\left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\mathrm{4}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}+…+\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}}\right)−\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}+\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }\:+\mathrm{1}−\mathrm{3}\sum_{\mathrm{k}=\mathrm{3}} ^{\mathrm{n}+\mathrm{2}} \:\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} } \\ $$$$=\mathrm{4}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}}\right)−\xi_{\mathrm{n}+\mathrm{1}} \left(\mathrm{2}\right)+\mathrm{1}−\mathrm{3}\left(\xi_{\mathrm{n}+\mathrm{2}} \left(\mathrm{2}\right)−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\mathrm{S}_{\mathrm{n}} =\mathrm{2}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}+\mathrm{1}−\mathrm{3}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\frac{\mathrm{5}}{\mathrm{4}}\right) \\ $$$$=\mathrm{3}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\frac{\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{6}}\:+\frac{\mathrm{15}}{\mathrm{4}}\:=\frac{\mathrm{27}}{\mathrm{4}}−\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{3}} \\ $$$$ \\ $$

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