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Question Number 33713 by abdo imad last updated on 22/Apr/18
find tbe value of Σ_(n=2) ^∞ ((n^2 −n+1)/((n−1)^2 (n+1)^2 )) .
$${find}\:{tbe}\:{value}\:{of}\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{{n}^{\mathrm{2}} −{n}+\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:. \\ $$
Commented by prof Abdo imad last updated on 26/Apr/18
let put S_n =Σ_(k=2) ^n ((n^2 −n+1)/((n−1)^2 (n+1)^2 )) let?drcompose F(x)= ((x^2 −x +1)/((x−1)^2 (x+1)^2 )) F(x)= (a/(x−1)) +(b/((x−1)^2 )) +(b/(x+1)) +(c/((x+1)^2 )) b=lim_(x→1) (x−1)^2 F(x)= (1/4) c =lim_(x→−1) (x+1)^2 F(x)= (3/4) ⇒ F(x)= (a/(x−1)) +(1/(4(x−1)^2 )) +(b/(x+1)) +(3/(4(x+1)^2 )) F(0)=1 = −a +(1/4)+b +(3/4) = 1+b−a ⇒b=a ⇒ F(x)= (a/(x−1)) + (1/(4(x−1)^2 )) +(a/(x+1)) +(3/(4(x+1)^2 )) =((2ax)/(x^2 −1)) +(1/(4(x−1)^2 )) + (3/(4(x+1)^2 )) lim_(x→+∞) xF(x)=0 = 2a ⇒a=0 ⇒ F(x)= (1/(4(x−1)^2 )) + (3/(4(x+1)^2 )) ⇒ S_n = (1/4) Σ_(k=2) ^n (1/((k−1)^2 )) +(3/4) Σ_(k=2) ^n (1/((k+1)^2 )) but Σ_(k=2) ^n (1/((k−1)^2 )) = Σ_(k=1) ^(n−1) (1/k^2 ) =ξ_(n−1) (2)→ξ(2)=(π^2 /6) Σ_(k=2) ^n (1/((k+1)^2 )) = Σ_(k=3) ^(n+1) (1/k^2 ) = Σ_(k=1) ^(n+1) (1/k^2 ) −1−(1/4) =ξ_(n+1) (2) −(3/4) →ξ(2)−(3/4) =(π^2 /6) −(3/4) lim_(n→+∞) S_n =(1/4)(π^2 /6) +(3/4)( (π^2 /6) −(3/4)) =(π^2 /(24)) + ((3π^2 )/(24)) −(9/(16)) = (π^2 /6) −(9/(16)) .
$${let}\:{put}\:{S}_{{n}} =\sum_{{k}=\mathrm{2}} ^{{n}} \:\:\frac{{n}^{\mathrm{2}} −{n}+\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:{let}?{drcompose} \\ $$$${F}\left({x}\right)=\:\frac{{x}^{\mathrm{2}} −{x}\:+\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${F}\left({x}\right)=\:\frac{{a}}{{x}−\mathrm{1}}\:+\frac{{b}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{b}}{{x}+\mathrm{1}}\:\:+\frac{{c}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${b}={lim}_{{x}\rightarrow\mathrm{1}} \left({x}−\mathrm{1}\right)^{\mathrm{2}} {F}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${c}\:={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} {F}\left({x}\right)=\:\frac{\mathrm{3}}{\mathrm{4}}\:\Rightarrow \\ $$$${F}\left({x}\right)=\:\frac{{a}}{{x}−\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{4}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{b}}{{x}+\mathrm{1}}\:+\frac{\mathrm{3}}{\mathrm{4}\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${F}\left(\mathrm{0}\right)=\mathrm{1}\:=\:−{a}\:+\frac{\mathrm{1}}{\mathrm{4}}+{b}\:+\frac{\mathrm{3}}{\mathrm{4}}\:=\:\mathrm{1}+{b}−{a}\:\Rightarrow{b}={a}\:\Rightarrow \\ $$$${F}\left({x}\right)=\:\frac{{a}}{{x}−\mathrm{1}}\:+\:\frac{\mathrm{1}}{\mathrm{4}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{a}}{{x}+\mathrm{1}}\:\:+\frac{\mathrm{3}}{\mathrm{4}\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}{ax}}{{x}^{\mathrm{2}} −\mathrm{1}}\:\:+\frac{\mathrm{1}}{\mathrm{4}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\:\frac{\mathrm{3}}{\mathrm{4}\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${lim}_{{x}\rightarrow+\infty} {xF}\left({x}\right)=\mathrm{0}\:=\:\mathrm{2}{a}\:\Rightarrow{a}=\mathrm{0}\:\Rightarrow \\ $$$${F}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{4}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\:\frac{\mathrm{3}}{\mathrm{4}\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${S}_{{n}} =\:\frac{\mathrm{1}}{\mathrm{4}}\:\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{\left({k}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\mathrm{3}}{\mathrm{4}}\:\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }\:{but} \\ $$$$\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{\left({k}−\mathrm{1}\right)^{\mathrm{2}} }\:=\:\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:=\xi_{{n}−\mathrm{1}} \left(\mathrm{2}\right)\rightarrow\xi\left(\mathrm{2}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\sum_{{k}=\mathrm{2}} ^{{n}} \:\:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }\:=\:\sum_{{k}=\mathrm{3}} ^{{n}+\mathrm{1}} \:\:\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:=\:\sum_{{k}=\mathrm{1}} ^{{n}+\mathrm{1}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$=\xi_{{n}+\mathrm{1}} \left(\mathrm{2}\right)\:−\frac{\mathrm{3}}{\mathrm{4}}\:\rightarrow\xi\left(\mathrm{2}\right)−\frac{\mathrm{3}}{\mathrm{4}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} =\frac{\mathrm{1}}{\mathrm{4}}\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:+\frac{\mathrm{3}}{\mathrm{4}}\left(\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:−\frac{\mathrm{3}}{\mathrm{4}}\right) \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{24}}\:+\:\frac{\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{24}}\:−\frac{\mathrm{9}}{\mathrm{16}}\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:−\frac{\mathrm{9}}{\mathrm{16}}\:\:. \\ $$$$ \\ $$
Commented by sma3l2996 last updated on 26/Apr/18
I think you have an error on lines 14 and 15 −1−(1/4)=((−5)/4)
$${I}\:{think}\:{you}\:{have}\:{an}\:{error}\:{on}\:{lines}\:\mathrm{14}\:{and}\:\mathrm{15} \\ $$$$−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}=\frac{−\mathrm{5}}{\mathrm{4}} \\ $$
Commented by prof Abdo imad last updated on 26/Apr/18
yes yes you are right i will correct the answer thank you so much sir sma3l....
$${yes}\:{yes}\:{you}\:{are}\:{right}\:{i}\:{will}\:{correct}\:{the}\:{answer} \\ $$$${thank}\:{you}\:{so}\:{much}\:{sir}\:{sma}\mathrm{3}{l}…. \\ $$
Commented by prof Abdo imad last updated on 26/Apr/18
error from line 15Σ_(k=2) ^n (1/((k+1)^2 )) =ξ_(n+1) (2)−(5/4) →_(n→+∞) ξ(2)−(5/4) =(π^2 /6) −(5/4) ⇒ lim_(n→+∞) S_n = (1/4) (π^2 /6) +(3/4)( (π^2 /6) −(5/4)) =(π^2 /(24)) +((3π^2 )/(24)) −((15)/(16)) = (π^2 /6) −((15)/(16)) ★lim S_n =(π^2 /6) −((15)/(16))★
$${error}\:{from}\:{line}\:\mathrm{15}\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }\:=\xi_{{n}+\mathrm{1}} \left(\mathrm{2}\right)−\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\rightarrow_{{n}\rightarrow+\infty} \:\xi\left(\mathrm{2}\right)−\frac{\mathrm{5}}{\mathrm{4}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:−\frac{\mathrm{5}}{\mathrm{4}}\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} =\:\frac{\mathrm{1}}{\mathrm{4}}\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:+\frac{\mathrm{3}}{\mathrm{4}}\left(\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:−\frac{\mathrm{5}}{\mathrm{4}}\right) \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{24}}\:+\frac{\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{24}}\:−\frac{\mathrm{15}}{\mathrm{16}}\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:−\frac{\mathrm{15}}{\mathrm{16}} \\ $$$$\bigstar{lim}\:{S}_{{n}} =\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:−\frac{\mathrm{15}}{\mathrm{16}}\bigstar \\ $$
Answered by sma3l2996 last updated on 23/Apr/18
A_n =Σ_(n=2) ^∞ ((n^2 −n+1)/((n−1)^2 (n+1)^2 ))=Σ_(n=2) ^∞ ((a/(n−1))+(b/((n−1)^2 ))+(c/(n+1))+(d/((n+1)^2 ))) with a=c=0 ; b=(1/4) ; d=(3/4) Σ_(n=2) ^∞ ((n^2 −n+1)/((n−1)^2 (n+1)^2 ))=(1/4)Σ_(n=2) ^∞ ((1/((n−1)^2 ))+(3/((n+1)^2 ))) =(1/4)Σ_(n=2) ^∞ (1/((n−1)^2 ))+(3/4)Σ_(n=2) ^∞ (1/((n+1)^2 )) let m=n−1 and k=n+1 so Σ_(n=2) ^∞ ((n^2 −n+1)/((n−1)^2 (n+1)^2 ))=(1/4)Σ_(m=1) ^∞ (1/m^2 )+(3/4)Σ_(k=3) ^∞ (1/k^2 ) =(1/4)ξ(2)+(3/4)(ξ(2)−((1/1^2 )+(1/2^2 )))=ξ(2)−(5/4) with (ξ(x) is zeta function) =(π^2 /6)−(5/4)
$${A}_{{n}} =\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{2}} −{n}+\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }=\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\left(\frac{{a}}{{n}−\mathrm{1}}+\frac{{b}}{\left({n}−\mathrm{1}\right)^{\mathrm{2}} }+\frac{{c}}{{n}+\mathrm{1}}+\frac{{d}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\right) \\ $$$${with}\:\:{a}={c}=\mathrm{0}\:\:;\:\:{b}=\frac{\mathrm{1}}{\mathrm{4}}\:\:;\:\:{d}=\frac{\mathrm{3}}{\mathrm{4}}\:\: \\ $$$$\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{2}} −{n}+\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{3}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{4}}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${let}\:\:{m}={n}−\mathrm{1}\:\:\:{and}\:\:\:{k}={n}+\mathrm{1} \\ $$$${so}\:\:\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{2}} −{n}+\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{m}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{4}}\underset{{k}=\mathrm{3}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\xi\left(\mathrm{2}\right)+\frac{\mathrm{3}}{\mathrm{4}}\left(\xi\left(\mathrm{2}\right)−\left(\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\right)\right)=\xi\left(\mathrm{2}\right)−\frac{\mathrm{5}}{\mathrm{4}}\:{with}\:\:\left(\xi\left({x}\right)\:\:{is}\:{zeta}\:{function}\right) \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\frac{\mathrm{5}}{\mathrm{4}} \\ $$
Commented by prof Abdo imad last updated on 26/Apr/18
you have commited a error of calculus sir sma3l..
$${you}\:{have}\:{commited}\:{a}\:{error}\:{of}\:{calculus}\:{sir}\:{sma}\mathrm{3}{l}.. \\ $$
Commented by sma3l2996 last updated on 26/Apr/18
(1/4)ξ(2)+(3/4)(ξ(2)−(5/4))=ξ(2)−((15)/(16)) =(π^2 /6)−((15)/(16))
$$\frac{\mathrm{1}}{\mathrm{4}}\xi\left(\mathrm{2}\right)+\frac{\mathrm{3}}{\mathrm{4}}\left(\xi\left(\mathrm{2}\right)−\frac{\mathrm{5}}{\mathrm{4}}\right)=\xi\left(\mathrm{2}\right)−\frac{\mathrm{15}}{\mathrm{16}} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\frac{\mathrm{15}}{\mathrm{16}} \\ $$

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