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How does one evaluate the limit: $$ \lim_{n \to \infty} \frac{1}{n}\sum\limits_{k=1}^{\lfloor{\frac{n}{2}\rfloor}} \cos\Bigl(\frac{k\pi}{n}\Bigr)$$

Yes, i recognize this as soon as i saw the problem: $$\int\limits_{0}^{1}f(x) \ dx = \lim_{n \to \infty} \frac{1}{n} \sum\limits_{k=1}^{n} f\Bigl(\frac{r}{n}\Bigr)$$ but the problem is there is $\lfloor{\frac{n}{2}\rfloor}$.

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    See, I told you things look better when you show what you've tried! +1 for that.2010-10-26
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    @J.M: If i know something, i shall definitely try. The other posts for which i haven't got anything is because i couldn't proceed at the problem even.2010-10-26
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    For what it's worth, the odd and even cases look to approach the same limit. (Evaluate numerically, for example, at $n = 99$ and $n = 100$.) Here's a hint for the solution: can you relate your sum to the sum $\sum_{k=1}^n cos(k\pi/n)$, which you know how to handle?2010-10-26

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Hint: Consider two separate limits. The limit of n even going to infinity, and the limit of n odd going to infinity. If they both converge to the same limit you are done. Both problems should be easy to solve using your observation.

The first one converges to $1/2\int_0^1 cos(\pi x/2) dx$ and the second converges to $\int_0^{1/2}cos(\pi x) dx$

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    @II-Bhima: Hey thanks! Shall work it out and let you know of any update!2010-10-26
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    @II-Bhima: They aren't equal. For $n$ even its coming out to be equal to $\frac{-\pi}{4}$ but for $n$ odd its 02010-10-26
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    Well, if thats the case, you have a sequence with two subsequences converging to different limits. So the sequence can't converge2010-10-26
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    I don't have time to use pen and paper, but I think both limits would converge to 1/pi.2010-10-26
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    @Bhima: I think $$\int\limits_{0}^{1} \cos{\pi{x}/2} \ dx = \int\limits_{0}^{\pi/2} \cos{t} \cdot \frac{\pi}{2} \ dt= - \frac{\pi}{2}$$2010-10-26
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    @Bhima: So multiplying by 1/2 gives $-\frac{\pi}{4}$.2010-10-26
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    You need to check that integration, when you do the substitution you have that $dy = pi/2 dx$ and so the integral becomes $\int_0^{\pi/2} cos(y)\frac{2}{\pi} dy$2010-10-26
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    @Bhima: I think i forgot! thanks2010-10-26
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You can probably also use the formula

$$\sum_{k=0}^{m} \cos (\phi + k \alpha) = \frac{\sin((m+1)\alpha/2)\cos(\phi + m\alpha/2)}{\sin \alpha/2}$$

Which can be derived using complex numbers, or in a more elementary fashion, using the fact that

$\displaystyle 2\cos(\phi + k \alpha) \sin(\alpha/2)$ $\displaystyle = \sin (\phi + \alpha(k+1/2)) - \sin(\phi + \alpha(k-1/2))$

and telescoping the sum.