Another Interesting Series

Well, most of you would have definitely proved that the alternating sum of reciprocal of odd numbers equals π4\frac{\pi}{4}.

Here's a more generalized case of the series

k=0(1)k2k+1(mk+n)(mk+mn)=πm2csc(nπm)\sum_{k=0}^\infty (-1)^{k}\frac{2k+1}{(mk+n)(mk+m-n)} = \frac{\pi}{m^2}\csc\left(\frac{n\pi}{m}\right)

Proof

It can be easily proved that

0xn11+xmdx=πmcsc(nπm)\int_0^\infty \frac{x^{n-1}}{1+x^m} \mathrm{d}x = \frac{\pi}{m} \csc\left(\frac{n\pi}{m}\right)

by making the substitution y=11+xmy=\frac{1}{1+x^m} and then the integral converts to Beta Function and the final result follows by using Euler's Reflection Formula Γ(z)Γ(1z)=πsinπz\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin \pi z}

Now, all that remains is to split the integral as

0xn11+xmdx=01xn11+xmdx+1xn11+xmdx\int_0^\infty \frac{x^{n-1}}{1+x^m} \mathrm{d}x = \int_0^1 \frac{x^{n-1}}{1+x^m} \mathrm{d}x + \int_1^\infty \frac{x^{n-1}}{1+x^m} \mathrm{d}x

The first term evaluates to 01xn11+xmdx=01xn1k=0(1)kxmk dx=k=0(1)kmk+n\int_0^1 \frac{x^{n-1}}{1+x^m} \mathrm{d}x = \int_0^1 x^{n-1}\sum_{k=0}^\infty (-1)^k x^{mk}\ \mathrm{d}x = \sum_{k=0}^\infty \frac{(-1)^k}{mk+n}

Similarly, the second term can be evaluated by making the following substitution : y=1xy=\frac{1}{x} 1xn11+xmdx=01ymn11+ymdy=k=0(1)kmk+mn\int_1^\infty \frac{x^{n-1}}{1+x^m} \mathrm{d}x = \int_0^1 \frac{y^{m-n-1}}{1+y^m} \mathrm{d}y = \sum_{k=0}^\infty \frac{(-1)^k}{mk+m-n}

And thus the result follows. \square

Interesting part is that, the base case when m=2,n=1m=2,n=1, you get the above mentioned well known series of alternating sum of reciprocal of odd numbers that is

k=0(1)k2k+1=π4\sum_{k=0}^\infty \frac{(-1)^{k}}{2k+1} = \frac{\pi}{4}

And I am sure, that you all must have got some other, perhaps more elegant, proof for the same. So, please do share it with all of us.

Note by Kishlaya Jaiswal
4 years, 9 months ago

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what an amazing beautiful proof

Syed Shahabudeen - 4 years, 9 months ago

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Thank You So Much.

Kishlaya Jaiswal - 4 years, 9 months ago

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