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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. –  J. M. Oct 26 '10 at 8:59
    
@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. –  anonymous Oct 26 '10 at 9:02
    
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? –  Michael Lugo Oct 26 '10 at 16:46

2 Answers 2

up vote 1 down vote accepted

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! –  anonymous Oct 26 '10 at 8:37
    
@II-Bhima: They aren't equal. For $n$ even its coming out to be equal to $\frac{-\pi}{4}$ but for $n$ odd its 0 –  anonymous Oct 26 '10 at 8:47
    
Well, if thats the case, you have a sequence with two subsequences converging to different limits. So the sequence can't converge –  Il-Bhima Oct 26 '10 at 8:58
    
I don't have time to use pen and paper, but I think both limits would converge to 1/pi. –  Il-Bhima Oct 26 '10 at 9:22
    
@Bhima: I think $$\int\limits_{0}^{1} \cos{\pi{x}/2} \ dx = \int\limits_{0}^{\pi/2} \cos{t} \cdot \frac{\pi}{2} \ dt= - \frac{\pi}{2}$$ –  anonymous Oct 26 '10 at 9:48

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.

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