# Prove that $sum_{n=1}^{infty }left ( frac {sin((2n-1)x)}{(2n-1)x)}right )^k frac{(-1)^{n-1}}{2n-1}=frac π 4$ for $0lt xlt frac pi {2k}$

Mathematics Asked by Paras on August 13, 2020

Question :- Prove that
$$sum_{ n =1}^{infty } left{frac{sinleft(left[2n – 1right]xright)} {left(2n – 1right)x}right}^{k} frac{left(-1right)^{n – 1}}{2n – 1} = frac{π}{4} qquadmbox{for}quad 0lt xlt frac{pi}{2k}$$
While reading some papers , I came across this series.Unfortunately I do not have any link to the website since I took screenshot of It few months back .

• The Author claims that the above series is true for
$$0lt xlt pi/left(2kright)$$ . However he does not provide any mathematical proof instead he calculates the sum for different $$x$$ and $$k$$ like for $$k = 100$$ and $$x = pi/200$$ the above sum up to $$50$$ terms is
$$0.78539 81633 97448 30961 55824$$
which is very close to $$pi/4$$.

• I verified It myself for $$k = 1$$.

• Actually the author is working on various variations of the
Gregory-Leibniz series and series of form
$$frac{sinleft(mathrm{f}left(xright)right)} {mathrm{g}left(xright)} quadmbox{and}quad frac{cosleft(mathrm{f}left(xright)right)}{mathrm{g}left(xright)}$$

• $${tt Mathematica}$$ evaluates the series in terms of Lerch transcendent $$Phi$$ function. I couldn’t find any way to prove the given series.

Thank you for your help !!.

We can prove this result by integrating the complex function $$f(z) = left(frac{sinleft(left[2z - 1right]xright)} {left(2z - 1right)x}right)^{k} , frac{pi csc (pi z)}{2z-1}$$ around a square contour with vertices at $$pm left(N+ frac{1}{2}right)+ i left(N+ frac{1}{2} right)$$, where $$N$$ is some positive integer.

(I'm exploiting the fact that the function $$pi csc (pi z)$$ has simple poles at the integers with residues that alternate between $$1$$ and $$-1$$.)

The condition $$0 < x le frac{pi}{2k}$$ ensures that the integral vanishes as $$N to infty$$ through the positive integers.

Basically what's happening is that the exponential growth of $$sin^{k}left((2z-1)xright)$$ as $$Im(z) to pm infty$$ is being neutralized by the exponential decay of $$csc (pi z)$$ as $$Im(z) to pm infty$$.

More specifically, the magnitude of $$sin^{k}left((2z-1)xright)$$ grows like a constant time $$e^{pm 2kxIm(z)}$$ as $$Im(z) to pm infty$$, while the magnitude of $$csc(pi z)$$ decays like a constant times $$e^{mp pi Im(z)}$$ as $$Im(z) to pm infty$$.

So if we integrate around the contour and then let $$N to infty$$, we get begin{align} lim_{N to infty} oint f(z) = 0 &= 2 pi i left( sum_{n=-infty}^{infty}operatorname{Res}left[f(z), n right] + operatorname{Res}left[f(z), frac{1}{2} right] right) \ &= small2 pi i left(sum_{n=-infty}^{infty}left(frac{sinleft(left[2n - 1right]xright)} {left(2n - 1right)x}right)^{k} frac{(-1)^n}{2n-1} + frac{pi}{2}lim_{z to frac{1}{2}} left(frac{sinleft(left[2z - 1right]xright)} {left(2z - 1right)x}right)^{k} csc(pi z) right) \ &=2 pi i left( sum_{n=-infty}^{infty}left(frac{sinleft(left[2n - 1right]xright)} {left(2n - 1right)x}right)^{k} frac{(-1)^n}{2n-1} + frac{pi}{2} (1)(1) right) \ &=2 pi i left(sum_{n=-infty}^{infty} left(frac{sinleft(left[2n - 1right]xright)} {left(2n - 1right)x}right)^{k} frac{(-1)^n}{2n-1} + frac{pi}{2} right) . end{align}

Therefore, $$sum_{n=-infty}^{infty}left(frac{sinleft(left[2n - 1right]xright)} {left(2n - 1right)x}right)^{k} frac{(-1)^{n-1}}{2n-1} = frac{pi}{2}.$$

But notice that $$sum_{n=-infty}^{0} left(frac{sinleft(left[2n - 1right]xright)} {left(2n - 1right)x}right)^{k} frac{(-1)^{n-1}}{2n-1}= sum_{n=1}^{infty}left(frac{sinleft(left[2n - 1right]xright)} {left(2n - 1right)x}right)^{k} frac{(-1)^{n-1}}{2n-1}.$$

The result then follows.

Correct answer by Random Variable on August 13, 2020

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