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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∑n=1∞3n2(2n+1)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 .
letS=∑n=1∞3n2(2n+1)2S=limn→+∞SnwithSn=∑k=1n3n2(2n+1)2letdecomposeF(x)=3x2(2x+1)2F(x)=ax+bx2+c2x+1+d(2x+1)2b=limx→0x2F(x)=3d=limx→−12(2x+1)2F(x)=12F(x)=ax+3x2+c2x+1+12(2x+1)2F(1)=13=a+3+c3+43⇒1=3a+9+c+4=3a+c+13⇒3a+c=−12F(−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=−12c=24⇒F(x)=−12x+3x2+242x+1+12(2x+1)2Sn=∑k=1nF(k)={−12∑k=1n1k+24∑k=1n12k+1}+{3∑k=1n1k2+12∑k=1n1(2k+1)2}=An+Bnbut3∑k=1∞1k2=3.π26=π22∑k=1∞1(2k+1)2=∑k=0∞1(2k+1)2−1=∑k=1∞1k2−14∑k=1∞1k2−1=34π26−1=π28−1limn→∞Bn=π22+12(π28−1)=π22+3π24−12=5π24−12alsowehave∑k=1n1k=Hn∑k=1n12k+1=∑k=0n12k+1−1=1+13+14+…+12n+1−1=1+12+13+14+….+12n+12n+1−12−14−…−12n−1=H2n+1−12Hn−1An=−12Hn+24(H2n+1−12Hn−1)=−12Hn+24H2n+1−12Hn−24=24(H2n+1−H2n)−24butH2n+1−Hn=ln(2n+1)+γ+o(1n)−ln(n)−γ−o(1n)=ln(2n+1n)+o(1n)→ln(2)(n→+∞)solimn→+∞An=24ln(2)−24solimn→+∞Sn=24ln(2)−24+5π24−12S=24ln(2)+5π24−36.

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