1

Evaluate:

$$\lim_{n \to \infty}\frac1n\sum_{k=0}^{[\frac{n}{2}]}\cos\left(\frac {k\pi}{n}\right)$$ Where $[\frac {n}{2}]$ denotes Largest Integer not exceeding $\frac{n}{2}$.

I thought much about this problem but couldn't found any way to handle this,So I need your help to handle this. Thanks in advance.

Empty
  • 13,012
Chiranjeev_Kumar
  • 3,061
  • 16
  • 30

1 Answers1

2

$\cos(x)$ is a concave function on the interval $\left[0,\frac{\pi}{2}\right]$, hence by the Hermite-Hadamard inequality:

$$\frac{1}{n}\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\cos\left(\frac{\pi k}{n}\right) = \frac{n}{1+\left\lfloor\frac{n}{2}\right\rfloor}\left(O\left(\frac{1}{n}\right)+\int_{0}^{\pi/2}\cos(x)\,dx\right)$$ so the wanted limit is just $\large\color{red}{2}$.

Jack D'Aurizio
  • 353,855
  • Sir Thanx for answer. As it is found as duplicate, can I delete my question? – Chiranjeev_Kumar Oct 23 '15 at 17:07
  • @Chiranjeev: The question is yours, sure you can. – Jack D'Aurizio Oct 23 '15 at 17:10
  • @JackD'Aurizio if we want to solve this like $\int_{0}^{\frac{1}{2}} \cos(x \pi) dx$, i got upper limit by multiplying $\lfloor \frac{n}{2} \rfloor \times \frac{1}{n}$ as n goes to infinity it we get 1/2, Just curious can we use this method? and the answer will be $\frac{1}{\pi}$ – TheStudent Mar 03 '22 at 12:10