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Question Number 45598 by maxmathsup by imad last updated on 14/Oct/18
calculate Σ_(n=1) ^∞ ((2n+1)/(n^2 (4n^2 −1)))
$${calculate}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\:\frac{\mathrm{2}{n}+\mathrm{1}}{{n}^{\mathrm{2}} \left(\mathrm{4}{n}^{\mathrm{2}} −\mathrm{1}\right)} \\ $$
Commented by maxmathsup by imad last updated on 15/Oct/18
let S_n =Σ_(k=1) ^n ((2k+1)/(k^2 (4k^2 −1))) we have S_n =Σ_(k=1) ^n ((2k+1)/(k^2 (2k+1)(2k−1))) =Σ_(k=1) ^n (1/(k^2 (2k−1))) let decompose F(x)=(1/(x^2 (2x−1))) F(x)=(a/x) +(b/x^2 ) +(c/(2x−1)) b =lim_(x→0) x^2 F(x) =−1 c =lim_(x→(1/2)) (2x−1)F(x) =4 ⇒F(x)=(a/x) −(1/x^2 ) +(4/(2x−1)) F(1)=1 =a−1 +4 =a+3 ⇒a=−2 ⇒F(x)=−(2/x) −(1/x^2 ) +(4/(2x−1)) ⇒ Σ_(k=1) ^n F(k) =−2 Σ_(k=1) ^n (1/k) −Σ_(k=1) ^n (1/k^2 ) +4 Σ_(k=1) ^n (1/(2k−1)) but Σ_(k=1) ^n (1/k) =H_n Σ_(k=1) ^n (1/k^2 ) =ξ_n (2) Σ_(k=1) ^n (1/(2k−1)) = 1 +(1/3) +(1/5) +...+(1/(2n−1)) =1+(1/2) +(1/3)+(1/4) +(1/5) +....+(1/(2n−1)) +(1/(2n)) −(1/2) (1+(1/2) +....+(1/n)) =H_(2n) −(1/2)H_n ⇒ S_n = −2 H_n −ξ_n (2)+4 H_(2n) −2H_n =4{ H_(2n) −H_n }−ξ_n (2) but H_(2n) =ln(2n)+γ +o((1/n)) and H_n =ln(n)+γ +o((1/n)) ⇒ H_(2n) −H_n =ln(((2n)/n))+o((1/n)) →ln(2)(n→+∞) also ξ_n (2)→ξ(2) =(π^2 /6) (n→+∞) ⇒ S=lim_(n→+∞) S_n = 4ln(2)−(π^2 /6) .
$${let}\:{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{2}{k}+\mathrm{1}}{{k}^{\mathrm{2}} \left(\mathrm{4}{k}^{\mathrm{2}} −\mathrm{1}\right)}\:\:\:\:{we}\:{have}\:\:{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{2}{k}+\mathrm{1}}{{k}^{\mathrm{2}} \left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{2}{k}−\mathrm{1}\right)} \\ $$$$=\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} \left(\mathrm{2}{k}−\mathrm{1}\right)}\:\:{let}\:{decompose}\:{F}\left({x}\right)=\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left(\mathrm{2}{x}−\mathrm{1}\right)} \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}}\:+\frac{{b}}{{x}^{\mathrm{2}} }\:+\frac{{c}}{\mathrm{2}{x}−\mathrm{1}} \\ $$$${b}\:={lim}_{{x}\rightarrow\mathrm{0}} {x}^{\mathrm{2}} {F}\left({x}\right)\:=−\mathrm{1} \\ $$$${c}\:={lim}_{{x}\rightarrow\frac{\mathrm{1}}{\mathrm{2}}} \:\:\left(\mathrm{2}{x}−\mathrm{1}\right){F}\left({x}\right)\:=\mathrm{4}\:\Rightarrow{F}\left({x}\right)=\frac{{a}}{{x}}\:−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\frac{\mathrm{4}}{\mathrm{2}{x}−\mathrm{1}} \\ $$$${F}\left(\mathrm{1}\right)=\mathrm{1}\:={a}−\mathrm{1}\:+\mathrm{4}\:={a}+\mathrm{3}\:\Rightarrow{a}=−\mathrm{2}\:\Rightarrow{F}\left({x}\right)=−\frac{\mathrm{2}}{{x}}\:−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\frac{\mathrm{4}}{\mathrm{2}{x}−\mathrm{1}}\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:{F}\left({k}\right)\:=−\mathrm{2}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:−\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:+\mathrm{4}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}\:{but} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:={H}_{{n}} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:=\xi_{{n}} \left(\mathrm{2}\right) \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}\:=\:\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{5}}\:+…+\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\: \\ $$$$=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{5}}\:+….+\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{2}{n}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:+….+\frac{\mathrm{1}}{{n}}\right) \\ $$$$={H}_{\mathrm{2}{n}} −\frac{\mathrm{1}}{\mathrm{2}}{H}_{{n}} \:\Rightarrow\:{S}_{{n}} =\:−\mathrm{2}\:{H}_{{n}} −\xi_{{n}} \left(\mathrm{2}\right)+\mathrm{4}\:{H}_{\mathrm{2}{n}} −\mathrm{2}{H}_{{n}} \\ $$$$=\mathrm{4}\left\{\:{H}_{\mathrm{2}{n}} −{H}_{{n}} \right\}−\xi_{{n}} \left(\mathrm{2}\right)\:\:{but} \\ $$$${H}_{\mathrm{2}{n}} \:={ln}\left(\mathrm{2}{n}\right)+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:\:{and}\:{H}_{{n}} ={ln}\left({n}\right)+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:\Rightarrow \\ $$$${H}_{\mathrm{2}{n}} −{H}_{{n}} ={ln}\left(\frac{\mathrm{2}{n}}{{n}}\right)+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:\rightarrow{ln}\left(\mathrm{2}\right)\left({n}\rightarrow+\infty\right)\:{also} \\ $$$$\xi_{{n}} \left(\mathrm{2}\right)\rightarrow\xi\left(\mathrm{2}\right)\:=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:\left({n}\rightarrow+\infty\right)\:\Rightarrow\:{S}={lim}_{{n}\rightarrow+\infty} {S}_{{n}} =\:\mathrm{4}{ln}\left(\mathrm{2}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:. \\ $$

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