I need to show that $$\sum_{x=1}^{n} \cos^2(x)$$ is bounded above. I know that there's a similar formula for $$\sum_{x=1}^{n} \sin(2x)$$ but I can't seem to find it anywhere.
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Not sure what you try to mean with your second expression since it's the same as the first. – Zubzub Dec 05 '17 at 08:02
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Sorry, the second sum was a typo. I've fixed it. – Dec 05 '17 at 08:05
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So do you know the second sum or looking for it? – Gevorg Hmayakyan Dec 05 '17 at 08:06
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@Gevorg Hmayakyan I know it. I'm trying to find the upper bound on the first sum. – Dec 05 '17 at 08:07
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@Brahadeesh It's a finite sum. – Dec 05 '17 at 08:08
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So the second sum equals $\frac{\sin(n) \sin(1 + n)}{\sin(1)}$ – Gevorg Hmayakyan Dec 05 '17 at 08:11
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The exact sum for the first one is $\frac{(2 n-1)}{4} + \frac{\sin(1 + 2 n)}{4\sin(1)}$ – Gevorg Hmayakyan Dec 05 '17 at 08:19
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Can you calculate this? – Gevorg Hmayakyan Dec 05 '17 at 08:23
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$$\begin{align}\sum_{k=1}^{n} \cos^2(k) &= \sum_{k=1}^n \frac{\cos(2k)+1}{2}=\frac 12(n+\sum_{k=1}^n\cos(2k))\\ &=\frac 12(n+Re(\sum_{k=1}^ne^{2ik})) = \frac 12(n+Re(e^{2i}\frac{e^{2in}-1}{e^{2i}-1}))\end{align}$$
Note that $\left|Re(e^{2i}\frac{e^{2in}-1}{e^{2i}-1})\right|\leq \frac{2}{|e^{2i}-1|}$, hence $$\sum_{k=1}^{n} \cos^2(k) = \frac n2 + O(1)$$
This implies the sum is not bounded.
Gabriel Romon
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