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Question Number 37341 by math khazana by abdo last updated on 12/Jun/18

calculate Σ_(n=1) ^∞ (3/(n^2 (2n+1)^2 ))

$${calculate}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\:\frac{\mathrm{3}}{{n}^{\mathrm{2}} \left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$

Commented by math khazana by abdo last updated on 14/Jun/18

let S=Σ_(n=1) ^∞ (3/(n^2 (2n+1)^2 )) S=lim_(n→+∞) S_n with S_n =Σ_(k=1) ^n (3/(n^2 (2n+1)^2 )) let decompose F(x)= (3/(x^2 (2x+1)^2 )) F(x)= (a/x) +(b/x^2 ) +(c/(2x+1)) +(d/((2x+1)^2 )) b=lim_(x→0) x^2 F(x)=3 d=lim_(x→−(1/2)) (2x+1)^2 F(x)= 12 F(x) = (a/x) +(3/x^2 ) +(c/(2x+1)) + ((12)/((2x+1)^2 )) F(1)=(1/3)= a +3 +(c/3) +(4/3) ⇒ 1=3a +9 +c +4 =3a+c +13 ⇒3a+c=−12 F(−1)= 3 =−a +3 −c +12 =−a−c +15⇒ −a−c=−12 ⇒a+c =12⇒c=12−a ⇒ 3a +12−a =−12 ⇒2a=−24 ⇒a =−12 c =24 ⇒ F(x) = ((−12)/x) +(3/x^2 ) +((24)/(2x+1)) +((12)/((2x+1)^2 )) S_n = Σ_(k=1) ^n F(k) ={−12 Σ_(k=1) ^n (1/k) +24Σ_(k=1) ^n (1/(2k+1))} +{3 Σ_(k=1) ^n (1/k^2 ) +12 Σ_(k=1) ^n (1/((2k+1)^2 ))}=A_n +B_n but3 Σ_(k=1) ^∞ (1/k^2 ) =3.(π^2 /6) =(π^2 /2) Σ_(k =1) ^∞ (1/((2k+1)^2 )) =Σ_(k=0) ^∞ (1/((2k+1)^2 )) −1 = Σ_(k=1) ^∞ (1/k^2 ) −(1/4)Σ_(k=1) ^∞ (1/k^2 )−1 =(3/4) (π^2 /6)−1 =(π^2 /8) −1 lim_(n→∞) B_n = (π^2 /2) +12((π^2 /8) −1) =(π^2 /2) +((3π^2 )/4) −12 =((5π^2 )/4) −12 also we have Σ_(k=1) ^n (1/k) =H_n Σ_(k=1) ^n (1/(2k+1)) =Σ_(k=0) ^n (1/(2k+1)) −1= 1 +(1/3) +(1/4) +...+(1/(2n+1))−1 =1 +(1/2) +(1/3) +(1/4) +.... +(1/(2n)) +(1/(2n+1)) −(1/2)−(1/4) −...−(1/(2n))−1 = H_(2n+1) −(1/2) H_n −1 A_n = −12 H_n +24( H_(2n+1) −(1/2)H_n −1) =−12H_n +24H_(2n+1) −12 H_n −24 =24( H_(2n+1) −H_(2n) ) −24 but H_(2n+1) −H_n =ln(2n+1) +γ +o((1/n))−ln(n)−γ −o((1/n)) =ln(((2n+1)/n)) +o((1/n))→ln(2)(n→+∞) so lim _(n→+∞) A_n =24 ln(2) −24 so lim_(n→+∞) S_n = 24ln(2)−24 +((5π^2 )/4) −12 S=24ln(2) +((5π^2 )/4) −36 .

$${let}\:{S}=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{3}}{{n}^{\mathrm{2}} \left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${S}={lim}_{{n}\rightarrow+\infty} {S}_{{n}} \:\:\:{with}\:{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{3}}{{n}^{\mathrm{2}} \left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${let}\:{decompose}\:{F}\left({x}\right)=\:\frac{\mathrm{3}}{{x}^{\mathrm{2}} \left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${F}\left({x}\right)=\:\frac{{a}}{{x}}\:+\frac{{b}}{{x}^{\mathrm{2}} }\:+\frac{{c}}{\mathrm{2}{x}+\mathrm{1}}\:+\frac{{d}}{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${b}={lim}_{{x}\rightarrow\mathrm{0}} {x}^{\mathrm{2}} {F}\left({x}\right)=\mathrm{3} \\ $$$${d}={lim}_{{x}\rightarrow−\frac{\mathrm{1}}{\mathrm{2}}} \:\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} {F}\left({x}\right)=\:\mathrm{12} \\ $$$${F}\left({x}\right)\:=\:\frac{{a}}{{x}}\:+\frac{\mathrm{3}}{{x}^{\mathrm{2}} }\:\:+\frac{{c}}{\mathrm{2}{x}+\mathrm{1}}\:+\:\frac{\mathrm{12}}{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${F}\left(\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{3}}=\:{a}\:+\mathrm{3}\:+\frac{{c}}{\mathrm{3}}\:\:+\frac{\mathrm{4}}{\mathrm{3}}\:\Rightarrow \\ $$$$\mathrm{1}=\mathrm{3}{a}\:+\mathrm{9}\:+{c}\:+\mathrm{4}\:=\mathrm{3}{a}+{c}\:+\mathrm{13}\:\Rightarrow\mathrm{3}{a}+{c}=−\mathrm{12} \\ $$$${F}\left(−\mathrm{1}\right)=\:\mathrm{3}\:=−{a}\:+\mathrm{3}\:−{c}\:+\mathrm{12}\:=−{a}−{c}\:+\mathrm{15}\Rightarrow \\ $$$$−{a}−{c}=−\mathrm{12}\:\Rightarrow{a}+{c}\:=\mathrm{12}\Rightarrow{c}=\mathrm{12}−{a}\:\Rightarrow \\ $$$$\mathrm{3}{a}\:+\mathrm{12}−{a}\:=−\mathrm{12}\:\Rightarrow\mathrm{2}{a}=−\mathrm{24}\:\Rightarrow{a}\:=−\mathrm{12} \\ $$$${c}\:=\mathrm{24}\:\Rightarrow \\ $$$${F}\left({x}\right)\:=\:\frac{−\mathrm{12}}{{x}}\:+\frac{\mathrm{3}}{{x}^{\mathrm{2}} }\:+\frac{\mathrm{24}}{\mathrm{2}{x}+\mathrm{1}}\:+\frac{\mathrm{12}}{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${S}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} {F}\left({k}\right)\:=\left\{−\mathrm{12}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:\:+\mathrm{24}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\right\} \\ $$$$+\left\{\mathrm{3}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:\:+\mathrm{12}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }\right\}={A}_{{n}} \:+{B}_{{n}} \\ $$$${but}\mathrm{3}\:\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:=\mathrm{3}.\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\sum_{{k}\:=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }\:=\sum_{{k}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }\:−\mathrm{1} \\ $$$$=\:\sum_{{k}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:−\frac{\mathrm{1}}{\mathrm{4}}\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }−\mathrm{1}\:=\frac{\mathrm{3}}{\mathrm{4}}\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\mathrm{1}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:−\mathrm{1} \\ $$$${lim}_{{n}\rightarrow\infty} \:{B}_{{n}} \:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\:+\mathrm{12}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:−\mathrm{1}\right)\:=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\:+\frac{\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{4}}\:−\mathrm{12} \\ $$$$=\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{4}}\:−\mathrm{12}\:\:{also}\:{we}\:{have} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:={H}_{{n}} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\:=\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\:−\mathrm{1}=\:\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{4}}\:+...+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}−\mathrm{1} \\ $$$$=\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{4}}\:+....\:+\frac{\mathrm{1}}{\mathrm{2}{n}}\:+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$−...−\frac{\mathrm{1}}{\mathrm{2}{n}}−\mathrm{1}\:=\:{H}_{\mathrm{2}{n}+\mathrm{1}} \:−\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{{n}} \:−\mathrm{1} \\ $$$${A}_{{n}} =\:−\mathrm{12}\:{H}_{{n}} \:\:+\mathrm{24}\left(\:{H}_{\mathrm{2}{n}+\mathrm{1}} \:−\frac{\mathrm{1}}{\mathrm{2}}{H}_{{n}} −\mathrm{1}\right) \\ $$$$=−\mathrm{12}{H}_{{n}} \:+\mathrm{24}{H}_{\mathrm{2}{n}+\mathrm{1}} \:−\mathrm{12}\:{H}_{{n}} \:−\mathrm{24} \\ $$$$=\mathrm{24}\left(\:{H}_{\mathrm{2}{n}+\mathrm{1}} \:−{H}_{\mathrm{2}{n}} \right)\:−\mathrm{24}\:\:{but} \\ $$$${H}_{\mathrm{2}{n}+\mathrm{1}} \:−{H}_{{n}} ={ln}\left(\mathrm{2}{n}+\mathrm{1}\right)\:+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)−{ln}\left({n}\right)−\gamma\:−{o}\left(\frac{\mathrm{1}}{{n}}\right) \\ $$$$={ln}\left(\frac{\mathrm{2}{n}+\mathrm{1}}{{n}}\right)\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\rightarrow{ln}\left(\mathrm{2}\right)\left({n}\rightarrow+\infty\right)\:{so} \\ $$$${lim}\:_{{n}\rightarrow+\infty} {A}_{{n}} \:\:=\mathrm{24}\:{ln}\left(\mathrm{2}\right)\:−\mathrm{24}\:{so} \\ $$$${lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} \:=\:\mathrm{24}{ln}\left(\mathrm{2}\right)−\mathrm{24}\:+\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{4}}\:−\mathrm{12} \\ $$$${S}=\mathrm{24}{ln}\left(\mathrm{2}\right)\:+\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{4}}\:−\mathrm{36}\:. \\ $$$$ \\ $$$$ \\ $$

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