Check the solution to "How could one prove that: ∑j=2∞∏k=1j2kj+k−1=π"

Jessie Jenkins

Answered question

2022-01-21

How could one prove that:

\sum_{j = 2}^{\infty} \prod_{k = 1}^{j} \frac{2 k}{j + k - 1} = π

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) = \frac{\arcsin x}{\sqrt{1 - x^{2}}}

f (x)

has a Maclurin expansion as follows:

f (x) = \sum_{n = 0}^{\infty} \frac{2^{2 n}}{(2 n + 1) (\begin{matrix} 2 n \\ n \end{matrix})} x^{2 n + 1}

Differentiating, we get

f^{'} (x) = \frac{x \arcsin x}{{(1 - x^{2})}^{\frac{3}{2}}} + \frac{1}{1 - x^{2}} = \sum_{n = 0}^{\infty} \frac{2^{2 n}}{(\begin{matrix} 2 n \\ n \end{matrix})} x^{2 n}

Evaluate at

x = \frac{1}{\sqrt{2}}

f^{'} (\frac{1}{\sqrt{2}}) = \frac{π}{2} + 2 = \sum_{n = 0}^{\infty} \frac{2^{n}}{(\begin{matrix} 2 n \\ n \end{matrix})}

Thus we have established that

\sum_{n = 2}^{\infty} \frac{2^{n}}{(\begin{matrix} 2 n \\ n \end{matrix})} = \frac{π}{2}

Now consider the original sum:

\sum_{n = 2}^{\infty} \prod_{k = 1}^{n} \frac{2 k}{n + k - 1} = \sum_{n = 2}^{\infty} \frac{2^{\cap}!}{n (n + 1) \dots (2 n - 1)}

= \sum_{n = 2}^{\infty} \frac{2^{\cap}! (n - 1)!}{(2 n - 1)!}

= 2 \sum_{n = 2}^{\infty} \frac{2^{n}}{(\begin{matrix} 2 n \\ n \end{matrix})}

= 2 \frac{π}{2}

= π

RizerMix

Expert2022-01-27Added 656 answers

Define the sequence b by

b_{0} = 2

and

b_{n} = \prod_{k = 1}^{n} \frac{2 k}{n + k - 1}

for

n \geq 1

, so we want to compute

\sum_{n = 2}^{\infty} b_{n}

. Let

a_{0}, a_{1}, a_{2}, . . .

be the sequence

4, - 4, - \frac{4}{3}, - \frac{4}{5}, - \frac{4}{7}, . . .

and let

△

be the difference operator on sequences defined by

(△ α)_{n} = α_{n + 1} - α_{n}

. Then one can check that the repeated difference at 0 equals

(△^{n} a)_{0} = (- 1)^{n} 2^{n + 1} b_{n}

By Eulers

nick1337

Expert2022-01-27Added 777 answers

Ill

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