# Infinite series involving 'x' has a constant value

• MHB
In summary, the sum of the first $n$ terms of the series is $1 - \frac1{2^n}\prod_{k=1}^n\cos^2\left(\frac{\pi x}{2^k}\right)$.

How to prove that $\sum_{i=1}^{\infty}\frac{1}{2^{3i}}\left(\csc^{2}\left(\frac{\pi x}{2^{i}}\right)+1\right)\sec^{2}\left(\frac{\pi x}{2^{i}}\right)\sin^{2}\left(\pi x\right)=1$ for all $$x\in\mathbb{R}$$.
Using graph, we can see that the value of this series is 1 for all values of x.
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Use the result $\sin x = 2\sin\frac x2\cos\frac x2$ repeatedly to see that \begin{aligned}\sin^2(\pi x) &= 2^2\sin^2\left(\frac{\pi x}2\right)\cos^2\left(\frac{\pi x}2\right) \\ &= 2^4\sin^2\left(\frac{\pi x}4\right)\cos^2\left(\frac{\pi x}4\right)\cos^2\left(\frac{\pi x}2\right) \\ & \vdots \\ &= 2^{2k}\sin^2\left(\frac{\pi x}{2^k}\right) \prod_{i=1}^k \cos^2\left(\frac{\pi x}{2^i}\right) .\end{aligned} You can then write the $k$th term of the series as $$\frac1{2^{3k}}\left(\csc^{2}\left(\frac{\pi x}{2^{k}}\right)+1\right)\sec^{2}\left(\frac{\pi x}{2^{k}}\right)\sin^{2}\left(\pi x\right) = \frac1{2^k} \left(1 + \sin^{2}\left(\frac{\pi x}{2^{k}}\right) \right)\prod_{i=1}^{k-1} \cos^2\left(\frac{\pi x}{2^i}\right) .$$ Using that, you should be able to show by induction that the sum of the first $n$ terms of the series is $$S_n(x) = \sum_{k=1}^{n}\frac{1}{2^{3k}}\left(\csc^{2}\left(\frac{\pi x}{2^{k}}\right)+1\right)\sec^{2}\left(\frac{\pi x}{2^{k}}\right)\sin^{2}\left(\pi x\right) = 1 - \frac1{2^n}\prod_{k=1}^n\cos^2\left(\frac{\pi x}{2^k}\right) .$$ That product of cosines lies between $0$ and $1$. It follows that $S_n(x)$ lies between $1 - \dfrac1{2^n}$ and $1$ for all $x$. Thus $S_n(x)$ converges (uniformly) to the constant function $1$.

Great. Thanks.
How did you find the value of $$S_{n}$$?

Satya said:
How did you find the value of $$S_{n}$$?
It's obvious from the graph at #1 that $S_n(x)$ increases very rapidly to $1$ as $n$ increases. So it seemed helpful to write $S_n(x)$ in the form $1$ - ?. I found that $$S_1(x) = 1 - \frac12\cos^2\left(\frac{\pi x}2\right),$$ $$S_2(x) = 1 - \frac14\cos^2\left(\frac{\pi x}4\right)\cos^2\left(\frac{\pi x}2\right),$$ $$S_3(x) = 1 - \frac18\cos^2\left(\frac{\pi x}8\right)\cos^2\left(\frac{\pi x}4\right)\cos^2\left(\frac{\pi x}2\right).$$ From there, it was easy enough to guess the general formula.