How could one prove that: \sum_{j=2}^\infty\prod_{k=1}^j\frac{2k}{j+k-1}=\pi

Jessie Jenkins

Jessie Jenkins

Answered question

2022-01-21

How could one prove that:
j=2k=1j2kj+k1=π

Answer & Explanation

Kingston Gates

Kingston Gates

Beginner2022-01-22Added 8 answers

This is going to be a little out of the blue, but here goes.
Consider the function
f(x)=arcsinx1x2
f(x) has a Maclurin expansion as follows:
f(x)=n=022n(2n+1)(2nn)x2n+1
Differentiating, we get
f(x)=xarcsinx(1x2)32+11x2=n=022n(2nn)x2n
Evaluate at x=12:
f(12)=π2+2=n=02n(2nn)
Thus we have established that
n=22n(2nn)=π2
Now consider the original sum:
n=2k=1n2kn+k1=n=22!n(n+1)(2n1)
=n=22!(n1)!(2n1)!
=2n=22n(2nn)
=2π2
=π
RizerMix

RizerMix

Expert2022-01-27Added 656 answers

Define the sequence b by b0=2 and bn=k=1n2kn+k1 for n1, so we want to compute n=2bn. Let a0,a1,a2,... be the sequence 4,4,43,45,47,... and let be the difference operator on sequences defined by (α)n=αn+1αn. Then one can check that the repeated difference at 0 equals (na)0=(1)n2n+1bn By Eulers
nick1337

nick1337

Expert2022-01-27Added 777 answers

Ill

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