Another Tricky Infinite Sum

Calculus Level 3

2 ⋅ ∑ n = 0 ∞ ( − 1 ) n ⋅ ( π / 5 ) 2 n ( 2 n ) ! = ? \large 2 \cdot \sum_{n=0}^{\infty} (-1)^{n} \cdot \frac{(\pi/5)^{2n}}{(2n)!}=\, ?

e e 3 \sqrt{3} π \pi ϕ \phi

Drex Beckman by Drex Beckman , 23, USA

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1 solution

Chew-Seong Cheong
Mar 11, 2016

2 ∑ n = 0 ∞ ( − 1 ) n ( π 5 ) 2 n ( 2 n ) ! = 2 cos ⁡ ( π 5 ) Maclaurin series = 2 ( 1 + 5 4 ) = φ See Note \begin{aligned} 2\sum_{n=0}^\infty \frac{(-1)^n \left(\frac{\pi}{5}\right)^{2n}}{(2n)!} & = 2\cos \left(\frac{\pi}{5} \right) \quad \quad \small \color{#3D99F6}{\text{Maclaurin series}} \\ & = 2\left(\color{#3D99F6}{\frac{1+\sqrt{5}}{4}} \right) = \boxed{\varphi} \quad \quad \small \color{#3D99F6}{\text{See Note}} \end{aligned}

Note: \color{#3D99F6}{\text{Note:}}

Using the identity:

cos ⁡ ( π 5 ) + cos ⁡ ( 3 π 5 ) = 1 2 cos ⁡ ( π 5 ) − cos ⁡ ( 2 π 5 ) = 1 2 cos ⁡ ( π 5 ) − 2 cos ⁡ 2 ( π 5 ) + 1 = 1 2 4 cos ⁡ 2 ( π 5 ) − 2 cos ⁡ ( π 5 ) − 1 = 0 ⇒ cos ⁡ ( π 5 ) = 2 ± 4 + 16 8 = 1 + 5 4 cos ⁡ ( π 5 ) > 0 \begin{aligned} \cos \left(\frac{\pi}{5} \right) + \cos \left(\frac{3\pi}{5} \right) & = \frac{1}{2} \\ \cos \left(\frac{\pi}{5} \right) - \cos \left(\frac{2\pi}{5} \right) & = \frac{1}{2} \\ \cos \left(\frac{\pi}{5} \right) - 2 \cos^2 \left(\frac{\pi}{5} \right) + 1 & = \frac{1}{2} \\ 4 \cos^2 \left(\frac{\pi}{5} \right) - 2 \cos \left(\frac{\pi}{5} \right) - 1& = 0 \\ \Rightarrow \cos \left(\frac{\pi}{5} \right) & = \frac{2 \pm \sqrt{4+16}}{8} \\ & = \frac{1+\sqrt{5}}{4} \quad \quad \small \color{#3D99F6}{\cos \left(\frac{\pi}{5} \right) > 0} \end{aligned}

11 Helpful 0 Interesting 0 Brilliant 1 Confused

How did the Maclaurin Series get there?

Oon Han - 2 years, 5 months ago

Do you mean how did I key it in? You can see the LaTex code by place your mouse cursor on top of the formula.

Chew-Seong Cheong - 2 years, 5 months ago

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No, I meant how did you evaluate the Maclaurin series to 2cos(pi / 5)?

Oon Han - 2 years, 5 months ago

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By Maclaurin serise cos ⁡ x = ∑ n = 0 ∞ ( − 1 ) n x 2 n ( 2 n ) ! \cos x = \displaystyle \sum_{n=0}^\infty \frac {(-1)^n x^{2n}}{(2n)!} . Putting x = π 5 x=\frac \pi 5 .

Chew-Seong Cheong - 2 years, 5 months ago

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