Trigonometric Infinite Series

∑n=0∞sin⁡(3π10(2n+1))−sin⁡(π10(2n+1))(2n+1)2=2G5\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: G=∑n=0∞(−1)n(2n+1)2≈0.916\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

#Calculus

Note by Ishan Singh
4 years, 11 months ago

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Comments

Let us consider sin⁡((2n+1)3π10)−sin⁡((2n+1)π10)\displaystyle \sin \left((2n+1)\frac{3\pi}{10}\right) - \sin \left((2n+1)\frac{\pi}{10}\right) for the first few nn.

n=0sin⁡3π10−sin⁡π10=cos⁡π5+cos⁡3π5=12n=1sin⁡π10−sin⁡3π10=−cos⁡3π5−cos⁡π5=−12n=2sin⁡3π2−sin⁡π2=−1−1=−2n=3sin⁡π10−sin⁡3π10==−12n=4sin⁡3π10−sin⁡π10==12n=5−sin⁡3π10+sin⁡π10==−12n=6−sin⁡π10+sin⁡3π10==12n=7sin⁡π2−sin⁡3π2=2n=8−sin⁡π10+sin⁡3π10==12n=9−sin⁡3π10+sin⁡π10==−12.........\begin{array}{c}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}

  ⟹  sin⁡((2n+1)3π10)−sin⁡((2n+1)π10)={If n mod 5≠2:{12if n is even−12if n is oddIf n mod 5=2:{2if n+35 is even−2if n+35 is odd\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:

S=∑n=0∞sin⁡((2n+1)3π10)−sin⁡((2n+1)π10)(2n+1)2=1212−1232−252−1272+1292−12112+12132+2152+12172−12192+12212−12232−2252−12272+...=12∑n=0∞(−1)n(2n+1)2−52∑n=0∞(−1)n(10n+5)2=12∑n=0∞(−1)n(2n+1)2−110∑n=0∞(−1)n(2n+1)2=(12−110)G=2G5\begin{aligned} 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{aligned}

Chew-Seong Cheong - 4 years, 11 months ago

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Nice way of using observation!

Aditya Kumar - 4 years, 11 months ago

Similar to hummus lucas numbers question 😉

Aman Rajput - 4 years, 11 months ago

is there any other solution

Liu Yang - 7 months ago
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