3
$\begingroup$

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 $\lfloor{\frac{n}{2}\rfloor}$.

  • 0
    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

2 Answers 2

1

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 $\int_0^{1/2}cos(\pi x) dx$

  • 0
    @Bhima: I think i forgot! thanks2010-10-26
1

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.