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S-n-1-6-n-n-1-2n-1-1-show-that-S-converges-2-give-the-value-of-S-




Question Number 131428 by pticantor last updated on 04/Feb/21
 S=Σ_(n≥1) (6/(n(n+1)(2n+1)))  1) show that S converges  2) give the value of S
S=n16n(n+1)(2n+1)1)showthatSconverges2)givethevalueofS
Answered by JDamian last updated on 04/Feb/21
S=Σ_(n≥1)   (1/((n(n+1)(2n+1))/6))  (1)  a_n =((n(n+1)(2n+1))/6) = 1^2 +2^2 +3^2 +∙∙∙+n^2   (1/a_n ) = (1/(1^2 +2^2 +∙∙∙+n^2 )) <(1/n^2 )    ∀n>1  Σ_(n≥1)  (1/a_n ) < Σ_(n≥1) (1/n^2 )=(π^2 /6)  S < (π^2 /6)
S=Σn11n(n+1)(2n+1)6(1)an=n(n+1)(2n+1)6=12+22+32++n21an=112+22++n2<1n2n>1n11an<n11n2=π26S<π26
Answered by Dwaipayan Shikari last updated on 04/Feb/21
Σ_(n≥1) (6/(n(2n+1)))−Σ_(n≥1) (6/((n+1)(2n+1)))  =Σ_(n=0) ^∞ (6/((n+1)(2n+3)))−(Σ_(n≥0) ^∞ (6/((n+1)(2n+1)))−Σ_(n≥0) ^1 (6/((n+1)(2n+1))))  =6(ψ((3/2))+ψ((1/2)))+6  =6(2−4log(2))+6=6(3−4log(2))   ψ((3/2))=2−2log(2)  ψ((1/2))=−2log(2)
n16n(2n+1)n16(n+1)(2n+1)=n=06(n+1)(2n+3)(n06(n+1)(2n+1)1n06(n+1)(2n+1))=6(ψ(32)+ψ(12))+6=6(24log(2))+6=6(34log(2))ψ(32)=22log(2)ψ(12)=2log(2)
Answered by mathmax by abdo last updated on 04/Feb/21
1) we have (6/(n(n+1)(2n+1)))∼(6/(2n^3 ))=(3/n^3 )and Σ (3/n^3 )cv ⇒S cv  2)(S/6)=lim_(n→+∞) w_n  with w_n =Σ_(k=1) ^n  (1/(k(k+1)(2k+1)))  let decompose F(x)=(1/(x(x+1)(2x+1))) ,F(x)=(a/x)+(b/(x+1))+(c/(2x+1))  a=1 ,b =(1/((−1)(−1)))=1 ,c =(1/((−(1/2))(1/2))) =−4 ⇒F(x)=(1/x)+(1/(x+1))−(4/(2x+1))  w_n =Σ_(k^ =1) ^n (1/k)+Σ_(k=1) ^n  (1/(k+1))−4 Σ_(k=1) ^n  (1/(2k+1)) but  Σ_(k=1) ^n  (1/k)=H_n ∼log(n)+γ  Σ_(k=1) ^n  (1/(k+1)) =Σ_(k=2) ^(n+1)  (1/k) =H_(n+1) −1 ∼log(n+1)+γ−1  Σ_(k=1) ^n  (1/(2k+1)) =(1/3)+(1/5)+...+(1/(2n+1)) =1+(1/2)+(1/3)+....+(1/(2n))+(1/(2n+1))  −1−(1/2)−(1/4)−....−(1/(2n))=H_(2n+1) −(1/2)H_n −1∼log(2n+1)+γ  −(1/2)(log(n)+γ)−1 =log(2n+1)−(1/2)log(n)+(γ/2)−1 ⇒  −4Σ_(k=1) ^n  (1/(2k+1))∼−4log(2n+1)+2logn−2γ+4 ⇒  W_n ∼log(n)+γ+log(n+1)+γ−1 −4log(2n+1)+2log(n)−2γ +4 ⇒  W_n ∼log(n^3 )+log(n+1)−4log(2n+1)+3 ⇒  W_n ∼log(((n^4  +n^3 )/((2n+1)^4 )))+3 ⇒W_n →3−4log(2) ⇒  S =6 W_∞ =6(3−4log(2))
1)wehave6n(n+1)(2n+1)62n3=3n3andΣ3n3cvScv2)S6=limn+wnwithwn=k=1n1k(k+1)(2k+1)letdecomposeF(x)=1x(x+1)(2x+1),F(x)=ax+bx+1+c2x+1a=1,b=1(1)(1)=1,c=1(12)12=4F(x)=1x+1x+142x+1wn=k=1n1k+k=1n1k+14k=1n12k+1butk=1n1k=Hnlog(n)+γk=1n1k+1=k=2n+11k=Hn+11log(n+1)+γ1k=1n12k+1=13+15++12n+1=1+12+13+.+12n+12n+111214.12n=H2n+112Hn1log(2n+1)+γ12(log(n)+γ)1=log(2n+1)12log(n)+γ214k=1n12k+14log(2n+1)+2logn2γ+4Wnlog(n)+γ+log(n+1)+γ14log(2n+1)+2log(n)2γ+4Wnlog(n3)+log(n+1)4log(2n+1)+3Wnlog(n4+n3(2n+1)4)+3Wn34log(2)S=6W=6(34log(2))
Commented by pticantor last updated on 08/Feb/21
thank you too much
thankyoutoomuch