Trigonometric Infinite Series

$\sum_{n=0}^{\infty} \dfrac{ \sin\left( \dfrac{3 \pi}{10} (2n+1)\right)-\sin\left( \dfrac{\pi}{10} (2n+1)\right)}{(2n+1)^2} = \dfrac{2 G}{5}$

Prove the equation above.

Notation: $$\displaystyle G = \sum_{n=0}^\infty \dfrac{ (-1)^n}{(2n+1)^2} \approx 0.916$$ denotes the Catalan's constant.

This is a part of the set Formidable Series and Integrals

Note by Ishan Singh
1 year, 11 months ago

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Let us consider $$\displaystyle \sin \left((2n+1)\frac{3\pi}{10}\right) - \sin \left((2n+1)\frac{\pi}{10}\right)$$ for the first few $$n$$.

$$\begin{array}{} n = 0 & \sin \frac{3\pi}{10} - \sin \frac{\pi}{10} = \cos \frac{\pi}{5} + \cos \frac{3\pi}{5} & = \frac 12 \\ n = 1 & \sin \frac{\pi}{10} - \sin \frac{3\pi}{10} = -\cos \frac{3\pi}{5} - \cos \frac{\pi}{5} & = - \frac 12 \\ n = 2 & \sin \frac{3\pi}{2} - \sin \frac{\pi}{2} = -1-1 & = -2 \\ n = 3 & \sin \frac{\pi}{10} - \sin \frac{3\pi}{10} = & = - \frac 12 \\ n = 4 & \sin \frac{3\pi}{10} - \sin \frac{\pi}{10} = & = \frac 12 \\ n = 5 & -\sin \frac{3\pi}{10} + \sin \frac{\pi}{10} = & = - \frac 12 \\ n = 6 & -\sin \frac{\pi}{10} + \sin \frac{3\pi}{10} = & = \frac 12 \\ n = 7 & \sin \frac{\pi}{2} - \sin \frac{3\pi}{2} & = 2 \\ n = 8 & -\sin \frac{\pi}{10} + \sin \frac{3\pi}{10} = & = \frac 12 \\ n = 9 & -\sin \frac{3\pi}{10} + \sin \frac{\pi}{10} = & = -\frac 12 \\ ... & ... & ... \end{array}$$

$$\implies \displaystyle \sin \left((2n+1)\frac{3\pi}{10}\right) - \sin \left((2n+1)\frac{\pi}{10}\right) = \begin{cases} \text{If } n \text{ mod } 5 \ne 2: & \begin{cases} \frac 12 & \text{if } n \text{ is even} \\ - \frac 12 & \text{if } n \text{ is odd} \end{cases} \\ \text{If } n \text{ mod } 5 = 2: & \begin{cases} 2 & \text{if } \frac {n+3}5 \text{ is even} \\ -2 & \text{if } \frac {n+3}5 \text{ is odd} \end{cases} \end{cases}$$

The sum is as follows:

\begin{align} S & = \sum_{n=0}^\infty \frac {\sin \left((2n+1)\frac{3\pi}{10}\right) - \sin \left((2n+1)\frac{\pi}{10}\right)}{(2n+1)^2} \\ & = \frac {\frac12}{1^2} - \frac {\frac12}{3^2} - \frac {2}{5^2} - \frac {\frac12}{7^2} + \frac {\frac12}{9^2} - \frac {\frac12}{11^2} + \frac {\frac12}{13^2} + \frac {2}{15^2} + \frac {\frac12}{17^2} - \frac {\frac12}{19^2} + \frac {\frac12}{21^2} - \frac {\frac12}{23^2} - \frac {2}{25^2} - \frac {\frac12}{27^2} + ... \\ & = \frac 12 \sum_{n=0}^\infty \frac {(-1)^n}{(2n+1)^2} - \frac 52 \sum_{n=0}^\infty \frac {(-1)^n}{(10n+5)^2} \\ & = \frac 12 \sum_{n=0}^\infty \frac {(-1)^n}{(2n+1)^2} - \frac 1{10} \sum_{n=0}^\infty \frac {(-1)^n}{(2n+1)^2} \\ & = \left(\frac 12 - \frac 1{10} \right) G = \boxed{\dfrac {2G}5} \end{align}

- 1 year, 11 months ago

Nice way of using observation!

- 1 year, 11 months ago

Similar to hummus lucas numbers question 😉

- 1 year, 11 months ago