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$\sum_{n=1}^{\infty} {\frac{1}{4^n \cos^2 (\frac{\pi}{2^{n+2}})}}$

How can I calculate this? Since there are $4^n$ and $\cos^2x$, I tried: $$\sum_{n=1}^{\infty} {\frac{1}{4^n \cos^2 (\frac{\pi}{2^{n+2}})}} = 4\sum_{n=1}^{\infty}{\frac{\sin^2{\frac{\pi}{4 \cdot 2^n}}}{4^{n}\sin^2{\frac{\pi}{4\cdot2^{n-1}}}}}$$ to use $2\sin x \cos x = \sin2x$

Albus Dumbledore
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John. P
  • 583

2 Answers2

1

Notice $$\begin{align}\frac{1}{\cos^2\frac{\theta}{2}} &= \frac{2}{1+\cos\theta} = 2\frac{1 - \cos\theta}{1-\cos^2\theta} = \frac{4 - 2(1+\cos\theta)}{1-\cos^2\theta}\\ &= \frac{4}{\sin^2\theta} - \frac{2}{1-\cos\theta} = \frac{4}{\sin^2\theta} - \frac{1}{\sin^2\frac{\theta}{2}} \end{align} $$ We have $$\begin{align} \sum_{n=1}^p \frac{1}{4^n\cos^2\frac{\pi}{2^{n+2}}} &= \sum_{n=1}^p \left[ \frac{1}{4^{n-1}\sin^2\frac{\pi}{2^{n+1}}} - \frac{1}{4^n\sin^2\frac{\pi}{2^{n+2}}} \right]\\ &=\frac{1}{4^{1-1}\sin^2\frac{\pi}{2^{1+1}}} - \frac{1}{4^p\sin^2\frac{\pi}{2^{p+2}}}\\ &= \frac{1}{\sin^2\frac{\pi}{4}} - \frac{\frac{16}{\pi^2}}{\left(\frac{2^{p+2}}{\pi}\sin\frac{\pi}{2^{p+2}}\right)^2} \end{align} $$ Since $\lim\limits_{x\to 0} \frac{\sin x}{x} = 1$, the denominator in last term tends to $1$ as $p \to \infty$, As a result,

$$\sum_{n=1}^\infty \frac{1}{4^n\cos^2\frac{\pi}{2^{n+2}}} = \lim_{p\to\infty} \sum_{n=1}^p \frac{1}{4^n\cos^2\frac{\pi}{2^{n+2}}} = 2 - \frac{16}{\pi^2} $$

achille hui
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Here’s a hint/general roadmap as to how I did it. I did not use your manipulation by the way, so focus on the original form of the problem you provided. Use the fact that $$\sin(x)=2\sin\bigl(\frac{x}{2}\bigr)\cos\bigl(\frac{x}{2}\bigr)$$. Now, keep rewriting the recurring sin term in this expression similar to how I just did, where I halved the original argument. This generalizes to the product representation $$\sin(x) = 2^n\cos\bigl(\frac{x}{2^n}\bigr)\sin\bigl(\frac{x}{2^n}\bigr)\prod_{k=1}^{n-1}\cos\bigl(\frac{x}{2^k}\bigr).$$ To use this you’re gonna need to reindex the sum to start at $n=2$. Rewrite the inside of the sum (namely the inside of the $\cos$ argument) to allow you to plug in a certain value of $x$. You’re gonna need to use this identity to turn the sum into a telescopic series that will eventually result in the answer. I got $2-\frac{\pi^2}{16}$.

MichaelCatliMath
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