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Question Number 125918 by I want to learn more last updated on 15/Dec/20

Σ_(n = − ∞) ^∞ (1/((n + (1/3))^3 ))

∑∞n=−∞1(n+13)3

Answered by mindispower last updated on 15/Dec/20

f(z)=((πcot(πz))/((z+(1/3))^3 )),over Re^(iθ) ,θ∈[0,2π];R→∞ lim_(R→∞) ∫_C f(z)dz=2iπRes(f,C)=0 ∣cot(πz)∣<M,∣z∣→0 (1/((z+(1/3))^3 ))→0 when ∣z∣→∞ pols of f are −(1/3),order 3 k∈Z order 1 Res(f,k)=(1/((k+(1/3))^3 )) Res(f,−(1/3))=(1/2)(∂/∂z^2 ).πcot(πz) =(1/2)[2π^3 (1+cot^2 (πz))cot(πz))∣z=−(1/3) =π^3 ((4/3))(−(1/( (√3)))) =((−4π^3 )/(3(√3))) Σ_(k∈Z) ^ Res(f,k)+Res(f,−(1/3))=0 ⇒Σres(f,k)=−Res(f,−(1/3))=((4π^3 )/(3(√3)))

f(z)=πcot(πz)(z+13)3,overReiθ,θ∈[0,2π];R→∞limR→∞∫Cf(z)dz=2iπRes(f,C)=0∣cot(πz)∣<M,∣z∣→01(z+13)3→0when∣z∣→∞polsoffare−13,order3k∈Zorder1Res(f,k)=1(k+13)3Res(f,−13)=12∂∂z2.πcot(πz)=12[2π3(1+cot2(πz))cot(πz))∣z=−13=π3(43)(−13)=−4π333∑k∈ZRes(f,k)+Res(f,−13)=0⇒Σres(f,k)=−Res(f,−13)=4π333

Commented by I want to learn more last updated on 15/Dec/20

Thanks sir. I appreciate.

Thankssir.Iappreciate.

Answered by mnjuly1970 last updated on 16/Dec/20

Commented by I want to learn more last updated on 16/Dec/20

Thanks sir, i appreciate. But sir, where can i learn summation like this. please. Tell me a book to download. please.

Thankssir,iappreciate.Butsir,wherecanilearnsummationlikethis.please.Tellmeabooktodownload.please.

Commented by mnjuly1970 last updated on 16/Dec/20

thank you so much master.. 1. advanced calculus ...{_(2.Ruel− chuchill) ^(1.M urray− spiegel) 2. complex analysis ..(Ruel− churchill)

thankyousomuchmaster..1.advancedcalculus...{2.Ruel−chuchill1.Murray−spiegel2.complexanalysis..(Ruel−churchill)

Commented by I want to learn more last updated on 20/Dec/20

Wow, thanks sir, i really appreciate sir.

Wow,thankssir,ireallyappreciatesir.

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