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Question Number 45240 by maxmathsup by imad last updated on 10/Oct/18

calculate Σ_(n=2) ^∞ ((2n+1)/(n^4 −n^2 ))

$${calculate}\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{\mathrm{2}{n}+\mathrm{1}}{{n}^{\mathrm{4}} −{n}^{\mathrm{2}} } \\ $$

Commented by maxmathsup by imad last updated on 11/Oct/18

let decompose F(x)=((2x+1)/(x^4 −x^2 )) =((2x+1)/(x^2 (x−1)(x+1))) ⇒ F(x)=(a/x) +(b/x^2 ) +(c/(x−1)) +(d/(x+1)) b =lim_(x→0) x^2 F(x) =−1 c =lim_(x→1) (x−1)F(x)=(3/2) d =lim_(x→−1) (x+1)F(x) =((−1)/(−2)) =(1/2) ⇒F(x)=(a/x) −(1/x^2 ) +(3/(2(x−1))) +(1/(2(x+1))) F(2) = (5/(4.3)) =(5/(12)) =(a/2) −(1/4) +(3/2) +(1/6) ⇒5 =6a−3 +18 +2 ⇒5=6a +17 ⇒ 6a=−12 ⇒a =−2 ⇒F(x)=−(2/x) −(1/x^2 ) +(3/(2(x−1))) +(1/(2(x+1))) let S_n =Σ_(k=2) ^∞ ((2k+1)/(k^4 −k^2 )) ⇒S_n =Σ_(k=2) ^n F(k) =−2 Σ_(k=2) ^n (1/k) −Σ_(k=2) ^n (1/k^2 ) +(3/2) Σ_(k=2) ^n (1/(k−1)) +(1/2)Σ_(k=2) ^n (1/(k+1)) but Σ_(k=2) ^n (1/k)=H_n −(3/2) Σ_(k=2) ^n =ξ_n (2)−1 Σ_(k=2) ^n (1/(k−1)) =Σ_(k=1) ^(n−1) (1/k) =H_(n−1) Σ_(k=2) ^n (1/(k+1)) =Σ_(k=3) ^(n+1) (1/k)=H_(n+1) −1−(1/2)−(1/3) =H_(n+1) −(3/2) −(1/3) =H_(n+1) −((11)/6) ⇒ S_n =−2H_n +3 +ξ_n (2)−1 +(3/2) H_(n−1) +(1/2) H_(n+1) −((11)/(12)) =−2H_n +(3/2) H_(n−1) +(1/2) H_(n+1) +ξ_n (2)+2−((11)/(12)) ⇒ S_n =−2(ln(n) +γ +o((1/n)))+(3/2)( ln(n−1)+γ +o((1/n)))+(1/2)(ln(n)+γ +o((1/n)))+ξ_n (2)+((13)/(12)) =−2ln(n) +(3/2)ln(n−1) +(1/2)ln(n) +o((1/n)) +ξ_n (2)+((13)/(12)) =ln((√((n−1)^3 ))(√n))−ln(n^2 ) +o((1/n))+ξ_n (2)+((13)/(12)) =ln{(((n−1)(√(n(n−1))))/n^2 )}+o((1/n))+ξ_n (2)+((13)/(12)) ⇒ lim_(n→+∞) S_n =ξ(2)+((13)/(12)) =(π^2 /6) +((13)/(12)) .

$${let}\:{decompose}\:{F}\left({x}\right)=\frac{\mathrm{2}{x}+\mathrm{1}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} }\:=\frac{\mathrm{2}{x}+\mathrm{1}}{{x}^{\mathrm{2}} \left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}}\:+\frac{{b}}{{x}^{\mathrm{2}} }\:+\frac{{c}}{{x}−\mathrm{1}}\:+\frac{{d}}{{x}+\mathrm{1}} \\ $$$${b}\:={lim}_{{x}\rightarrow\mathrm{0}} {x}^{\mathrm{2}} {F}\left({x}\right)\:=−\mathrm{1} \\ $$$${c}\:={lim}_{{x}\rightarrow\mathrm{1}} \left({x}−\mathrm{1}\right){F}\left({x}\right)=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${d}\:={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right){F}\left({x}\right)\:=\frac{−\mathrm{1}}{−\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{F}\left({x}\right)=\frac{{a}}{{x}}\:−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\frac{\mathrm{3}}{\mathrm{2}\left({x}−\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{1}\right)} \\ $$$${F}\left(\mathrm{2}\right)\:=\:\frac{\mathrm{5}}{\mathrm{4}.\mathrm{3}}\:=\frac{\mathrm{5}}{\mathrm{12}}\:=\frac{{a}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{4}}\:+\frac{\mathrm{3}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{6}}\:\:\Rightarrow\mathrm{5}\:=\mathrm{6}{a}−\mathrm{3}\:+\mathrm{18}\:+\mathrm{2}\:\Rightarrow\mathrm{5}=\mathrm{6}{a}\:+\mathrm{17}\:\Rightarrow \\ $$$$\mathrm{6}{a}=−\mathrm{12}\:\Rightarrow{a}\:=−\mathrm{2}\:\Rightarrow{F}\left({x}\right)=−\frac{\mathrm{2}}{{x}}\:−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\frac{\mathrm{3}}{\mathrm{2}\left({x}−\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{1}\right)} \\ $$$${let}\:{S}_{{n}} =\sum_{{k}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{2}{k}+\mathrm{1}}{{k}^{\mathrm{4}} −{k}^{\mathrm{2}} }\:\Rightarrow{S}_{{n}} =\sum_{{k}=\mathrm{2}} ^{{n}} \:{F}\left({k}\right) \\ $$$$=−\mathrm{2}\:\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:−\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:+\frac{\mathrm{3}}{\mathrm{2}}\:\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{{k}−\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{1}}\:{but} \\ $$$$\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{{k}}={H}_{{n}} −\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\sum_{{k}=\mathrm{2}} ^{{n}} \:=\xi_{{n}} \left(\mathrm{2}\right)−\mathrm{1} \\ $$$$\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{{k}−\mathrm{1}}\:=\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\frac{\mathrm{1}}{{k}}\:={H}_{{n}−\mathrm{1}} \\ $$$$\sum_{{k}=\mathrm{2}} ^{{n}} \:\:\frac{\mathrm{1}}{{k}+\mathrm{1}}\:=\sum_{{k}=\mathrm{3}} ^{{n}+\mathrm{1}} \:\frac{\mathrm{1}}{{k}}={H}_{{n}+\mathrm{1}} \:−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}\:={H}_{{n}+\mathrm{1}} −\frac{\mathrm{3}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{3}}\:={H}_{{n}+\mathrm{1}} −\frac{\mathrm{11}}{\mathrm{6}}\:\Rightarrow \\ $$$${S}_{{n}} =−\mathrm{2}{H}_{{n}} \:\:+\mathrm{3}\:\:+\xi_{{n}} \left(\mathrm{2}\right)−\mathrm{1}\:+\frac{\mathrm{3}}{\mathrm{2}}\:{H}_{{n}−\mathrm{1}} \:+\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{{n}+\mathrm{1}} −\frac{\mathrm{11}}{\mathrm{12}} \\ $$$$=−\mathrm{2}{H}_{{n}} \:+\frac{\mathrm{3}}{\mathrm{2}}\:{H}_{{n}−\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{{n}+\mathrm{1}} \:+\xi_{{n}} \left(\mathrm{2}\right)+\mathrm{2}−\frac{\mathrm{11}}{\mathrm{12}}\:\Rightarrow \\ $$$${S}_{{n}} \:=−\mathrm{2}\left({ln}\left({n}\right)\:+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\right)+\frac{\mathrm{3}}{\mathrm{2}}\left(\:{ln}\left({n}−\mathrm{1}\right)+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\right)+\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left({n}\right)+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\right)+\xi_{{n}} \left(\mathrm{2}\right)+\frac{\mathrm{13}}{\mathrm{12}} \\ $$$$=−\mathrm{2}{ln}\left({n}\right)\:+\frac{\mathrm{3}}{\mathrm{2}}{ln}\left({n}−\mathrm{1}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({n}\right)\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:+\xi_{{n}} \left(\mathrm{2}\right)+\frac{\mathrm{13}}{\mathrm{12}} \\ $$$$={ln}\left(\sqrt{\left({n}−\mathrm{1}\right)^{\mathrm{3}} }\sqrt{{n}}\right)−{ln}\left({n}^{\mathrm{2}} \right)\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)+\xi_{{n}} \left(\mathrm{2}\right)+\frac{\mathrm{13}}{\mathrm{12}} \\ $$$$={ln}\left\{\frac{\left({n}−\mathrm{1}\right)\sqrt{{n}\left({n}−\mathrm{1}\right)}}{{n}^{\mathrm{2}} }\right\}+{o}\left(\frac{\mathrm{1}}{{n}}\right)+\xi_{{n}} \left(\mathrm{2}\right)+\frac{\mathrm{13}}{\mathrm{12}}\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} =\xi\left(\mathrm{2}\right)+\frac{\mathrm{13}}{\mathrm{12}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:+\frac{\mathrm{13}}{\mathrm{12}}\:. \\ $$

Answered by ajfour last updated on 11/Oct/18

Σ_(n=2) ^∞ ((2n+1)/(n^2 (n−1)(n+1))) =Σ(((n+1)+(n−1))/((n−1)nn(n+1)))+(1/2)Σ(((n+1)−(n−1))/((n−1)nn(n+1))) =Σ(1/((n−1)n^2 ))+Σ(1/(n^2 (n+1))) +(1/2)Σ(1/((n−1)n^2 ))−(1/2)Σ(1/(n^2 (n+1))) =(3/2)Σ(1/((n−1)n^2 ))+(1/2)Σ(1/(n^2 (n+1))) =(3/4)Σ((1/((n−1)n))−(1/n^2 ))+(1/4)Σ((1/n^2 )−(1/(n(n+1)))) =(3/4)Σ((1/(n−1))−(1/n))−(1/4)Σ((1/n)−(1/(n+1))) −(1/2)Σ(1/n^2 ) =(3/4)(1−(1/2)+(1/2)−(1/3)+..)−(1/4)((1/2)−(1/3)+(1/3)−(1/4)+..) −(1/2)((1/4)+(1/9)+...) = (3/4)−(1/8)−(1/2)((1/4)+(1/9)+(1/(16))+..) ln (1−x)=−(x+(x^2 /2)+(x^3 /3)+...) ∫_0 ^( 1) ((ln (1−x))/x)dx=−(1+(1/4)+(1/9)+(1/(16))+..) so Σ_(n=2) ^∞ ((2n+1)/(n^4 −n^2 )) = (3/4)−(1/8)−(1/2)[1+∫_0 ^( 1) ((ln (1−x))/x)dx] ..... (cant solve the integral, sir ).

$$\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\:\:\frac{\mathrm{2}{n}+\mathrm{1}}{{n}^{\mathrm{2}} \left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)} \\ $$$$=\Sigma\frac{\left({n}+\mathrm{1}\right)+\left({n}−\mathrm{1}\right)}{\left({n}−\mathrm{1}\right){nn}\left({n}+\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}}\Sigma\frac{\left({n}+\mathrm{1}\right)−\left({n}−\mathrm{1}\right)}{\left({n}−\mathrm{1}\right){nn}\left({n}+\mathrm{1}\right)} \\ $$$$=\Sigma\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right){n}^{\mathrm{2}} }+\Sigma\frac{\mathrm{1}}{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:+\frac{\mathrm{1}}{\mathrm{2}}\Sigma\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right){n}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}}\Sigma\frac{\mathrm{1}}{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\Sigma\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right){n}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}}\Sigma\frac{\mathrm{1}}{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}\Sigma\left(\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right){n}}−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)+\frac{\mathrm{1}}{\mathrm{4}}\Sigma\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}\Sigma\left(\frac{\mathrm{1}}{{n}−\mathrm{1}}−\frac{\mathrm{1}}{{n}}\right)−\frac{\mathrm{1}}{\mathrm{4}}\Sigma\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}\Sigma\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}+..\right)−\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}+..\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{9}}+...\right) \\ $$$$\:=\:\frac{\mathrm{3}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{16}}+..\right) \\ $$$$\mathrm{ln}\:\left(\mathrm{1}−{x}\right)=−\left({x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+...\right) \\ $$$$\int_{\mathrm{0}} ^{\:\:\mathrm{1}} \:\frac{\mathrm{ln}\:\left(\mathrm{1}−{x}\right)}{{x}}{dx}=−\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{16}}+..\right) \\ $$$${so}\: \\ $$$$\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\:\:\frac{\mathrm{2}{n}+\mathrm{1}}{{n}^{\mathrm{4}} −{n}^{\mathrm{2}} }\:=\:\frac{\mathrm{3}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}+\int_{\mathrm{0}} ^{\:\:\mathrm{1}} \frac{\mathrm{ln}\:\left(\mathrm{1}−{x}\right)}{{x}}{dx}\right] \\ $$$$.....\:\left({cant}\:{solve}\:{the}\:{integral},\:{sir}\:\right). \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18

((2n+1)/(n^2 (n+1)(n−1)))=(a/n)+(b/n^2 )+(c/(n+1))+(d/(n−1)) 2n+1=an(n+1)(n−1)+b(n+1)(n−1)+cn^2 (n−1)+dn^2 (n+1) put n=0 1=b(−1) b=−1 put n−1=0 3=d×1×2 d=(3/2) put n+1=0 −2+1=c×(−1)^2 ×(−2) −1=−2c c=(1/2) 2n+1=an(n+1)(n−1)+b(n+1)(n−1)+cn^2 (n−1)+dn^2 (n+1) 2n+1=a(n^3 −n)+(−1)(n^2 −1)+(1/2)n^2 (n−1)+(3/2)n^2 (n+1) 2n+1=n^3 (a+(1/2)+(3/2))+n^2 (−1−(1/2)+(3/2))+n(−a)+1 a+2=0 a=−2 =Σ_(n=2) ^∞ (((−2)/n)+((−1)/n^2 )+((1/2)/(n+1))+((3/2)/(n−1))) =S_1 +S_2 +S_3 +S_4 S_1 =−2Σ_(n=2) ^∞ (1/n) S_1 =−2{((1/1)+(1/2)+(1/3)+...∞)−(1/1)} =−2(γ−1)=−2γ+1 γ=Eulers constant S_2 =−1Σ_(n=2) ^∞ (1/n^2 ) =−1{((1/1^2 )+(1/2^2 )+(1/3^2 )+..∞)−(1/1)} =−((π^2 /6)−1)=1−(π^2 /6) S_3 =(1/2)Σ_(n=2) ^∞ (1/(n+1)) =(1/2){((1/1)+(1/2)+(1/3)+...∞)−((1/1)+(1/2))} =(1/2)(γ−(3/2))=(γ/2)−(3/4) S_4 =(3/2)Σ_(n=2) ^∞ (1/(n−1)) (3/2)((1/1)+(1/2)+(1/3)+...) =((3γ)/2) so reauired ans is S_1 +S_2 +S_3 +S_4 =(−2γ+1)+(1−(π^2 /6))+((γ/2)−(3/4))+(((3γ)/2)) =−2γ+2γ+1+1−(3/4)−(π^2 /6) =(5/4)−(π^2 /6) pls check...

$$\frac{\mathrm{2}{n}+\mathrm{1}}{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)\left({n}−\mathrm{1}\right)}=\frac{{a}}{{n}}+\frac{{b}}{{n}^{\mathrm{2}} }+\frac{{c}}{{n}+\mathrm{1}}+\frac{{d}}{{n}−\mathrm{1}} \\ $$$$\mathrm{2}{n}+\mathrm{1}={an}\left({n}+\mathrm{1}\right)\left({n}−\mathrm{1}\right)+{b}\left({n}+\mathrm{1}\right)\left({n}−\mathrm{1}\right)+{cn}^{\mathrm{2}} \left({n}−\mathrm{1}\right)+{dn}^{\mathrm{2}} \left({n}+\mathrm{1}\right) \\ $$$${put}\:{n}=\mathrm{0}\: \\ $$$$\mathrm{1}={b}\left(−\mathrm{1}\right)\:\:{b}=−\mathrm{1} \\ $$$${put}\:{n}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{3}={d}×\mathrm{1}×\mathrm{2}\:\:\:{d}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${put}\:{n}+\mathrm{1}=\mathrm{0} \\ $$$$−\mathrm{2}+\mathrm{1}={c}×\left(−\mathrm{1}\right)^{\mathrm{2}} ×\left(−\mathrm{2}\right) \\ $$$$−\mathrm{1}=−\mathrm{2}{c}\:\:\:{c}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{2}{n}+\mathrm{1}={an}\left({n}+\mathrm{1}\right)\left({n}−\mathrm{1}\right)+{b}\left({n}+\mathrm{1}\right)\left({n}−\mathrm{1}\right)+{cn}^{\mathrm{2}} \left({n}−\mathrm{1}\right)+{dn}^{\mathrm{2}} \left({n}+\mathrm{1}\right) \\ $$$$\mathrm{2}{n}+\mathrm{1}={a}\left({n}^{\mathrm{3}} −{n}\right)+\left(−\mathrm{1}\right)\left({n}^{\mathrm{2}} −\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}{n}^{\mathrm{2}} \left({n}−\mathrm{1}\right)+\frac{\mathrm{3}}{\mathrm{2}}{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right) \\ $$$$\mathrm{2}{n}+\mathrm{1}={n}^{\mathrm{3}} \left({a}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}\right)+{n}^{\mathrm{2}} \left(−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}\right)+{n}\left(−{a}\right)+\mathrm{1} \\ $$$${a}+\mathrm{2}=\mathrm{0}\:\:{a}=−\mathrm{2} \\ $$$$=\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\left(\frac{−\mathrm{2}}{{n}}+\frac{−\mathrm{1}}{{n}^{\mathrm{2}} }+\frac{\frac{\mathrm{1}}{\mathrm{2}}}{{n}+\mathrm{1}}+\frac{\frac{\mathrm{3}}{\mathrm{2}}}{{n}−\mathrm{1}}\right) \\ $$$$={S}_{\mathrm{1}} +{S}_{\mathrm{2}} +{S}_{\mathrm{3}} +{S}_{\mathrm{4}} \\ $$$${S}_{\mathrm{1}} =−\mathrm{2}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}} \\ $$$${S}_{\mathrm{1}} =−\mathrm{2}\left\{\left(\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+...\infty\right)−\frac{\mathrm{1}}{\mathrm{1}}\right\} \\ $$$$\:\:\:\:\:=−\mathrm{2}\left(\gamma−\mathrm{1}\right)=−\mathrm{2}\gamma+\mathrm{1}\:\:\gamma={Eulers}\:{constant} \\ $$$${S}_{\mathrm{2}} =−\mathrm{1}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:=−\mathrm{1}\left\{\left(\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+..\infty\right)−\frac{\mathrm{1}}{\mathrm{1}}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=−\left(\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\mathrm{1}\right)=\mathrm{1}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$${S}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}+\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\left(\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+...\infty\right)−\left(\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right\} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\gamma−\frac{\mathrm{3}}{\mathrm{2}}\right)=\frac{\gamma}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${S}_{\mathrm{4}} =\frac{\mathrm{3}}{\mathrm{2}}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}−\mathrm{1}} \\ $$$$\:\:\:\:\:\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+...\right) \\ $$$$\:\:\:\:\:=\frac{\mathrm{3}\gamma}{\mathrm{2}} \\ $$$${so}\:{reauired}\:{ans}\:{is}\:{S}_{\mathrm{1}} +{S}_{\mathrm{2}} +{S}_{\mathrm{3}} +{S}_{\mathrm{4}} \\ $$$$=\left(−\mathrm{2}\gamma+\mathrm{1}\right)+\left(\mathrm{1}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\right)+\left(\frac{\gamma}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{4}}\right)+\left(\frac{\mathrm{3}\gamma}{\mathrm{2}}\right) \\ $$$$=−\mathrm{2}\gamma+\mathrm{2}\gamma+\mathrm{1}+\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$=\frac{\mathrm{5}}{\mathrm{4}}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:\:\:{pls}\:{check}... \\ $$$$ \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18

Commented by ajfour last updated on 11/Oct/18

sir please explain me how Σ_(n=1) ^∞ (1/n^2 ) = (π^2 /6) .

$${sir}\:{please}\:{explain}\:{me}\:{how} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18

proof taken from SL Loney Trigonometry

$${proof}\:{taken}\:{from}\:{SL}\:{Loney}\:{Trigonometry} \\ $$

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