An easy sum

Geometry Level 2

What is the value of:

∑ r = 0 ∞ sin ⁡ 4 ( 2 r − 2 π ) 4 r ? \large \sum_{r=0}^\infty \frac{\sin^4 (2^{r-2}\pi)}{4^r}?

2 2 \frac{\sqrt{2}}{2} 1 4 \frac{1}{4} 1 2 \frac{1}{2} The series diverges 1 1

Russell Ngo by Russell Ngo , 26, United Kingdom

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1 solution

Chew-Seong Cheong
Aug 28, 2018

S = ∑ r = 0 ∞ sin ⁡ 4 ( 2 r − 2 π ) 4 r Note that sin ⁡ ( 2 r − 2 π ) = 0 for r ≥ 2 = ∑ r = 0 1 sin ⁡ 4 ( 2 r − 2 π ) 4 r = sin ⁡ 4 π 4 1 + sin ⁡ 4 π 2 4 = 1 4 + 1 4 = 1 2 \begin{aligned} S & = \sum_{r=0}^\infty \frac {\sin^4(2^{r-2}\pi)}{4^r} & \small \color{#3D99F6} \text{Note that }\sin (2^{r-2}\pi) = 0 \text{ for }r \ge 2 \\ & = \sum_{r=0}^1 \frac {\sin^4(2^{r-2}\pi)}{4^r} \\ & = \frac {\sin^4 \pi 4}{1} + \frac {\sin^4 \pi 2}{4} \\ & = \frac 14 + \frac 14 \\ & = \boxed{\dfrac 12} \end{aligned}

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