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

Here's a more generalized case of the series

\[\sum_{k=0}^\infty (-1)^{k}\frac{2k+1}{(mk+n)(mk+m-n)} = \frac{\pi}{m^2}\csc\left(\frac{n\pi}{m}\right)\]

**Proof**

It can be easily proved that

\[\int_0^\infty \frac{x^{n-1}}{1+x^m} \mathrm{d}x = \frac{\pi}{m} \csc\left(\frac{n\pi}{m}\right)\]

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

Now, all that remains is to split the integral as

\[\int_0^\infty \frac{x^{n-1}}{1+x^m} \mathrm{d}x = \int_0^1 \frac{x^{n-1}}{1+x^m} \mathrm{d}x + \int_1^\infty \frac{x^{n-1}}{1+x^m} \mathrm{d}x\]

The first term evaluates to \[\int_0^1 \frac{x^{n-1}}{1+x^m} \mathrm{d}x = \int_0^1 x^{n-1}\sum_{k=0}^\infty (-1)^k x^{mk}\ \mathrm{d}x = \sum_{k=0}^\infty \frac{(-1)^k}{mk+n}\]

Similarly, the second term can be evaluated by making the following substitution : \(y=\frac{1}{x}\) \[\int_1^\infty \frac{x^{n-1}}{1+x^m} \mathrm{d}x = \int_0^1 \frac{y^{m-n-1}}{1+y^m} \mathrm{d}y = \sum_{k=0}^\infty \frac{(-1)^k}{mk+m-n}\]

And thus the result follows. \(\square\)

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

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

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.

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TopNewestwhat an amazing beautiful proof

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Thank You So Much.

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