Talk:Pi as Sum of Alternating Sequence of Products of 3 Consecutive Reciprocals

I would like to give a simple proof of this result, but moving and creating pages is difficult and I am out of practice. Here is an outline:

Step 1: Partial Fraction Decomposition:

$\ds \frac 1 {(2n)(2n+1)(2n+2)} = \frac 1 2 \paren{\frac 1 {2n} - \frac 2 {2 n + 1} + \frac 1 {2 n + 2}}$

Step 2: Telescoping Series:

$\ds \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1}} {(2n)(2n+1)(2n+2)} = \cdots = \frac 1 4 + \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^n} {2 n + 1} = -\frac 3 4 + \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {2 n + 1}$

Step 3: Leibniz's Formula for Pi:

$\ds above = -\frac 3 4 + \frac \pi 4$

hence the result.

I am happy to fill in the rest when the bare bones of the relevant pages are created.

As for the page title, this sum is actually an alternating sum. Besides, it may be confused with Sum of Sequence of Products of 3 Consecutive Reciprocals.

--RandomUndergrad (talk) 12:09, 7 July 2021 (UTC)

There you go. What do you think abut the new title? Does it work for you? --prime mover (talk) 13:03, 7 July 2021 (UTC)

Yes, it works. Everything is done now. Thank you. --RandomUndergrad (talk) 14:10, 7 July 2021 (UTC)