Pi as Sum of Alternating Sequence of Products of 3 Consecutive Reciprocals/Proof 2

Theorem

$\dfrac {\pi - 3} 4 = \dfrac 1 {2 \times 3 \times 4} - \dfrac 1 {4 \times 5 \times 6} + \dfrac 1 {6 \times 7 \times 8} \cdots$

Proof

The alternating sum can be written as $\ds \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1}} {2 n \paren{2 n + 1} \paren{2 n + 2}}$.

By Partial Fraction Decomposition:

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

Therefore:

\(\ds \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } {2 n \paren{2 n + 1} \paren{2 n + 2} }\)

\(=\)

\(\ds \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } 2 \paren{\frac 1 {2 n} - \frac 2 {2 n + 1} + \frac 1 {2 n + 2} }\)

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

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

Translation of Index Variable of Summation

\(\ds \)

\(=\)

\(\ds \frac 1 4 + \frac \pi 4 - 1\)

Leibniz's Formula for Pi

\(\ds \)

\(=\)

\(\ds \frac {\pi - 3} 4\)

$\blacksquare$