Question Number 103879 by abony1303 last updated on 18/Jul/20

Commented byabony1303 last updated on 18/Jul/20

Please, help

Commented bymyear last updated on 18/Jul/20

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Answered by Ar Brandon last updated on 18/Jul/20

(n/(2n^3 +1))=(n/(((2)^(1/3) n+1)((4)^(1/3) n^2 −(2)^(1/3) n+1)))                =(a/((2)^(1/3) n+1))+((bn+c)/((4)^(1/3) n^2 −(2)^(1/3) n+1))                =((a((4)^(1/3) n^2 −(2)^(1/3) n+1)+(bn+c)((2)^(1/3) n+1))/(((2)^(1/3) n+1)((4)^(1/3) n^2 −(2)^(1/3) n+1)))  (4)^(1/3) a+(2)^(1/3) b=0 , a+c=0 , −(2)^(1/3) a+b+(2)^(1/3) c=1  ⇒b−2(2)^(1/3) a=1 ⇒ 3(4)^(1/3) a=−(2)^(1/3) ⇒a=((−1)/(3(2)^(1/3) )) , c=(1/(3(2)^(1/3) )) , b=(1/3)  ⇒(n/(2n^3 +1))=((−1/3(2)^(1/3) )/((2)^(1/3) n+1))+((n/3+1/3(2)^(1/3) )/((4)^(1/3) n^2 −(2)^(1/3) n+1))  Σ_(n=1) ^∞ (n/(2n^3 +1))=Σ_(n=1) ^∞ {((n/3+1/3(2)^(1/3) )/((4)^(1/3) n^2 −(2)^(1/3) n+1))−((1/3(2)^(1/3) )/((2)^(1/3) n+1))}    Any idea to proceed ?

Answered by ~blr237~ last updated on 19/Jul/20

Let state  f(z)=(z/(z^3 +a^3 ))  g(z)=(π/(tan(πz)))     a>0  P(fg)={−a,aw_0 ,aw_1 ,−n,n /    n∈N ,w_0 =e^(i(π/3)) .....}  let state   points A(−in),B(−n+in),C(n+in) D(in)  for n≥0  δ_n =ABCDis a rectangle (close way) and   f(z)=O((1/(∣z∣^2 )))  (•)  The residus theorems allows us to say  ∫_δ_n  f(z)g(z)dz= 2πi Σ_(x∈P(fg)) ind(δ_n ,x)Res(fg,x)=2πi(Σ_(k=0) ^n f(k) +Res(fg, aw_0 )+Res(fg,aw_1 )]   because   (•) prove that  the first term →0 when n→∞   So  S=Σ_(n=0) ^∞ f(n)= −[((aw_0 g(aw_0 ))/(3(aw_0 )^2 )) +((aw_1 g(aw_1 ))/(3(aw_1 )^2 ))]  S=−(2/3) Re(((aw_0 g(aw_0 ))/((aw_0 )^2 )))   cause  w_(1 ) =w_0 ^−    and  g(z^− )=g^− (z)  S=−((2π)/(3a)) Re((1/(   w_0 tan(πaw_0 ))))         tan(πw_0 )=  −i ((e^(i2aπw_0 ) −1)/(e^(i2πaw_0 ) +1))   and 2πaiw_0 =2iπa((1/2)+i((√3)/2))=−π(√3) +iπa  tan(πw_0 )=−i ((e^(iπa) e^(−π(√3)) +1)/(e^(iπa) e^(−π(√3)) −1))=−i((e^(−2π(√3)) −1−2ie^(−π(√3)) sin(πa))/(e^(−2π(√3)) +1−2cos(πa)e^(−π(√3)) ))  S=((πe^(π(√3)) )/(6a(e^(−2π(√3)) +1−2e^(−π(√3)) cos(πa))))   Re( (((√3)+i)/(sh(π(√3))+isin(πa))))  S=((πe^(2π(√3)) )/(12a(ch(π(√3))−cos(πa))) ×(((√3) sh(π(√3))−sin(πa))/(sh^2 (π(√3))+sin^2 (πa)))         Finally                                         S(a)= Σ_(n=1) ^∞   (n/(n^3 + a^3 )) = ((π(√3)e^(2π(√3)) (sh(π(√3))−sin(πa)))/(12a(ch(π(√3))−cos(πa))(sh^2 (π(√3))+sin^2 (πa))))      a>0    Let   deduce your by    (1/2) S(^3 (√2)) .