Prove that : ∑k=0∞(−1)k(2k+1)3=π332

Question

Prove that :  ∑k=0∞(−1)k(2k+1)3=π332

Darian Sosa · Accepted Answer

You can evaluate this sum using the residue theorem. First, note that it may be extended out to −∞:  ∑k=0∞(−1)k(2k+1)3=12∑k=−∞∞(−1)k(2k+1)3  From the residue theorem:  ∑k=−∞∞(−1)k(2k+1)3=−Resz=−12πcsc⁡πz(2z+1)3=−18Resz=−12πcsc⁡πz(z+12)3  This residue involves taking the second derivative of the csc term. Note that, for a generic function f(z) having a triple pole at z=z0:  Resz=z0f(z)=12!limz→z0d2dz2[(z−z0)3f(z)]  so that  Resz=−12πcsc⁡πz(z+12)3=π32![csc⁡πzcot2πz+csc3πz]z=−12=−π32  Putting this all together, we get the stated result:  ∑k=0∞(−1)k(2k+1)3=π332

Flickkorbma · Accepted Answer

I&#039;ll show you a related series and then pick a special case. We&#039;ll start with ∑k≥1ekiθk=−log⁡(1=eiθ) and we&#039;ll take the imaginary part of both sides to get ∫0β∫0α∑k≥1sin⁡(kθ)kdθdα=∫0β∑0β∑k≥1kβ−sin⁡(kβ)k3 =∫0β∫0απ2−θ2dθdα=14(πβ2−β33)=β2(3π−β)12 This way, we see that ∑k≥1sin⁡(kβ)k3=β3−3πβ2+2π2β12 Now, just pick β=π2 and ∑k≥0(−1)k(2k+1)3=−∑k≥1(−1)k(2k−1)3=π332 To fix some issues in convergence, I&#039;ll take my first equation and introduce a new variable 0&amp;lt;t&amp;lt;1 ∑k≥tkekiθk=−log⁡(1−teiθ) and then take the imaginary parts and let t→1: limt→1[∑k≥1tksin⁡(kθ)k]=limt→1[tan−1(tsin⁡(θ)tcos⁡(θ)−1)]=π2−θ2 and we proceed normally from here.

RizerMix · Accepted Answer

The Polylogarithm function is defined as Lis(z)=∑k=1∞zkks Now Li3(i)=∑k=1∞ikks and Li3(−i)=∑k=1∞(−1)kikks Hence, Li3(i)−Li3(−i)=2i(∑k=0∞(−1)k(2k+1)3) (1) Now the PolyLogarithmic function satisfies a very nice identity Lin(e2πix)+(−1)nLin(e−2πix)=−2πi)nn!Bn(x) Taking n=3 and x=14, gives us Li3(i)−Lin(−i)=−(2πi)33!B3(1/4)=i8π36364=iπ316 (2) Comparing (1) and (2) gives us (∑k=0∞(−1)k(2k+1)3)=π332

Prove that : \sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^3}=\frac{\pi^3}{32}

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