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

  • ${tt Mathematica}$ 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 !!.

One Answer

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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