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
k=0∑∞(−1)k(mk+n)(mk+m−n)2k+1=m2πcsc(mnπ)
Proof
It can be easily proved that
∫0∞1+xmxn−1dx=mπcsc(mnπ)
by making the substitution y=1+xm1 and then the integral converts to Beta Function and the final result follows by using Euler's Reflection Formula Γ(z)Γ(1−z)=sinπzπ
Now, all that remains is to split the integral as
∫0∞1+xmxn−1dx=∫011+xmxn−1dx+∫1∞1+xmxn−1dx
The first term evaluates to
∫011+xmxn−1dx=∫01xn−1k=0∑∞(−1)kxmk dx=k=0∑∞mk+n(−1)k
Similarly, the second term can be evaluated by making the following substitution : y=x1
∫1∞1+xmxn−1dx=∫011+ymym−n−1dy=k=0∑∞mk+m−n(−1)k
And thus the result follows. □
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
k=0∑∞2k+1(−1)k=4π
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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Top Newestwhat an amazing beautiful proof
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Thank You So Much.
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