Power of Four and Sine Summation

Geometry Level 5

∑ n = 1 ∞ 4 n sin ⁡ 4 ( π 2 n ) \large\displaystyle\sum_{n=1}^{\infty}4^n\sin^4\left(\dfrac{\pi}{2^n}\right)

If the value of the summation above is in the form of π a b \dfrac{\pi^a}b , where a a and b b are positive integers, find the value of a + b a+b .


The answer is 3.

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Ikkyu San by Ikkyu San , 23, Thailand

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1 solution

Chew-Seong Cheong
Apr 27, 2016

Using the identity below:

sin ⁡ 2 θ = 2 2 sin ⁡ 2 θ 2 cos ⁡ 2 θ 2 sin ⁡ 2 θ = 4 sin ⁡ 2 θ 2 ( 1 − sin ⁡ 2 θ 2 ) ⟹ sin ⁡ 2 θ + 4 sin ⁡ 4 θ 2 = 4 sin ⁡ 2 θ 2 \begin{aligned} \sin^2 \theta & = 2^2 \sin^2 \frac \theta 2 \cos^2 \frac \theta 2 \\ \sin^2 \theta & = 4 \sin^2 \frac \theta 2 (1 - \sin^2 \frac \theta 2) \\ \implies \sin^2 \theta + 4 \sin^4 \frac \theta 2 & = 4 \sin^2 \frac \theta 2 \end{aligned}

Let a m = 4 m sin ⁡ 4 π 2 m a_m = 4^m \sin^4 \dfrac{\pi}{2^m} , S m = ∑ n = 1 m a n S_m = \displaystyle \sum_{n=1}^m a_n and θ = π 2 m \theta = \dfrac{\pi}{2^m} . Then we have:

\(\begin{array} {} m = 1 & \implies \sin^2 \pi + 4 \sin^4 \frac \pi 2 = 4 \sin^2 \frac \pi 2 & \implies S_1 = a_1 = 4 \sin^2 \frac \pi 2 \\ m = 2 & \implies 4 \sin^2 \frac \pi 2 + 4^2 \sin^4 \frac \pi 4 = 4^2 \sin^2 \frac \pi 4 & \implies S_2 = a_1 + a_2 = 4^2 \sin^2 \frac \pi 2 \end{array} \)

We note that S m = 4 m sin ⁡ 2 ( π 2 m ) S_m = 4^m \sin^2 \left(\frac{\pi}{2^m} \right) for m = 1 m = 1 and 2 2 . Let claim it is true for all m m and prove it by induction.

  1. For m = 1 m=1 , S 1 = 4 sin ⁡ 4 ( π 2 ) = 4 = 4 sin ⁡ 2 ( π 2 ) S_1 = 4 \sin ^4 \left(\frac{\pi}{2} \right) = 4 = 4 \sin^2 \left(\frac{\pi}{2} \right) . The claim is true for m = 1 m = 1 .
  2. Assume that the claim is true for m m , then

S m + 1 = S m + 4 m + 1 sin ⁡ 4 ( π 2 m + 1 ) = 4 m sin ⁡ 2 ( π 2 m ) + 4 m + 1 sin ⁡ 4 ( π 2 m + 1 ) = 4 m ( 2 sin ⁡ ( π 2 m + 1 ) cos ⁡ ( π 2 m + 1 ) ) 2 + 4 m + 1 sin ⁡ 4 ( π 2 m + 1 ) = 4 m + 1 sin ⁡ 2 ( π 2 m + 1 ) ( cos ⁡ 2 ( π 2 m + 1 ) + sin ⁡ 2 ( π 2 m + 1 ) ) ⇒ S m + 1 = 4 m + 1 sin ⁡ 2 ( π 2 m + 1 ) \begin{aligned} \quad \quad S_{m+1} & = S_m + 4^{m+1} \sin^4 \left(\frac{\pi}{2^{m+1}} \right) \\ & = 4^m \sin^2 \left(\frac{\pi}{2^m} \right) + 4^{m+1} \sin^4 \left(\frac{\pi}{2^{m+1}} \right) \\ & = 4^m \left(2 \sin \left(\frac{\pi}{2^{m+1}} \right) \cos \left(\frac{\pi}{2^{m+1}} \right) \right)^2+ 4^{m+1} \sin^4 \left(\frac{\pi}{2^{m+1}} \right) \\ & = 4^{m+1} \sin^2 \left(\frac{\pi}{2^{m+1}} \right) \left(\cos^2 \left(\frac{\pi}{2^{m+1}} \right) + \sin^2 \left(\frac{\pi}{2^{m+1}} \right) \right) \\ \Rightarrow S_{m+1} & = 4^{m+1} \sin^2 \left(\frac{\pi}{2^{m+1}} \right) \end{aligned}

\quad \quad The claim is also true for m + 1 m+1 and hence true for all m m .

Therefore, we have:

lim ⁡ m → ∞ S m = lim ⁡ m → ∞ 4 m + 1 sin ⁡ 2 ( π 2 m + 1 ) = lim ⁡ m → ∞ ( sin ⁡ ( π 2 m + 1 ) π 2 m + 1 ) 2 π 2 = π 2 \begin{aligned} \lim_{m \to \infty} S_m & = \lim_{m \to \infty} 4^{m+1} \sin^2 \left(\frac{\pi}{2^{m+1}} \right) \\ & = \lim_{m \to \infty} \left( \frac{\sin \left(\frac{\pi}{2^{m+1}} \right)}{\frac{\pi}{2^{m+1}}}\right)^2 \pi^2 \\ & = \pi^2 \end{aligned}

⇒ a + b = 2 + 1 = 3 \Rightarrow a + b = 2 + 1 = \boxed{3}

13 Helpful 0 Interesting 0 Brilliant 0 Confused

Can you please tell how you arrived at what you claimed?

Mayank Chaturvedi - 5 years, 1 month ago

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Thanks for asking. I have added a few lines to explain it.

Chew-Seong Cheong - 5 years, 1 month ago

What motivated the claim?

A Former Brilliant Member - 5 years, 1 month ago

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What motivated the claim was of course to find a general form so that we can solve the problem by putting m → ∞ m \to \infty . How did I get the general form is as the few lines I have added.

Chew-Seong Cheong - 5 years, 1 month ago

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Thanks for the explaination. (By the way, manier times, it is difficult the find the closed form of the partial sums. This was the reason I asked for the motivation. I did it in this way as I had already encountered the closed form of the partial sums elsewhere.)

A Former Brilliant Member - 5 years, 1 month ago

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@A Former Brilliant Member – If you did not see/guess the closed form, it can be derived directly. The sum to N N terms is ∑ n = 1 N 4 n − 1 [ 1 − cos ⁡ π 2 n − 1 ] 2 = 3 2 ∑ n = 0 N − 1 4 n − 2 ∑ n = 0 N − 1 4 n cos ⁡ π 2 n + 2 ∑ n = − 1 N − 2 4 n cos ⁡ π 2 n \sum_{n=1}^N 4^{n-1}\big[1 - \cos\tfrac{\pi}{2^{n-1}}\big]^2 \; = \; \tfrac32\sum_{n=0}^{N-1}4^n - 2\sum_{n=0}^{N-1} 4^n \cos\tfrac{\pi}{2^n} + 2\sum_{n=-1}^{N-2}4^n\cos\tfrac{\pi}{2^n} which is easy to evaluate, being a GP and a telescoping series.

Mark Hennings - 5 years, 1 month ago

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@Mark Hennings – Thanks for this method!

A Former Brilliant Member - 5 years, 1 month ago

You may shorten the proof by removing those lines, since this is already shown at beginning.

= 4 m ( 2 sin ⁡ ( π 2 m + 1 ) cos ⁡ ( π 2 m + 1 ) ) 2 + 4 m + 1 sin ⁡ 4 ( π 2 m + 1 ) = 4 m + 1 sin ⁡ 2 ( π 2 m + 1 ) ( cos ⁡ 2 ( π 2 m + 1 ) + sin ⁡ 2 ( π 2 m + 1 ) ) \begin{aligned} & = 4^m \left(2 \sin \left(\frac{\pi}{2^{m+1}} \right) \cos \left(\frac{\pi}{2^{m+1}} \right) \right)^2+ 4^{m+1} \sin^4 \left(\frac{\pi}{2^{m+1}} \right) \\ & = 4^{m+1} \sin^2 \left(\frac{\pi}{2^{m+1}} \right) \left(\cos^2 \left(\frac{\pi}{2^{m+1}} \right) + \sin^2 \left(\frac{\pi}{2^{m+1}} \right) \right) \end{aligned}

Abdelhamid Saadi - 5 years, 1 month ago

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