The limit of a sequence involving cosine How do I show that
$$\lim_{N\to\infty} \frac{1}{2N+1} \sum_{n=-N}^{N} \left( \frac{1+\cos\left(\frac{\pi}{2}n \right)}{2}\right) = \frac{1}{2}?$$
Can anyone please show the steps of this limit ?
 A: Cosine is an even function, and so is $1+cos(x)$. So if we take out the term when $n=0$, then we can start the series at $n=1$, and the expression simplifies to
$$\lim_{N\to \infty} \frac{1}{2N+1} \sum_{n=1}^{N}\bigg(1+\cos\bigg(\frac{\pi}{2}n\bigg)\bigg) +\frac{1}{2(2N+1)}$$
$$=\lim_{N\to \infty} \frac{1}{2N+1} \bigg(\sum_{n=1}^{N}\bigg(\cos\bigg(\frac{\pi}{2}n\bigg)\bigg)+\sum_{n=1}^{N}1\bigg) +\frac{1}{2(2N+1)}$$
$$=\lim_{N\to \infty} \frac{1}{2N+1} \sum_{n=1}^{N}\bigg(\cos\bigg(\frac{\pi}{2}n\bigg)\bigg)+\frac{N}{2N+1}+\frac{1}{2(2N+1)}$$
$$= 0+\frac{1}{2}+0=\frac{1}{2}$$
The first term goes to $0$ since it's a telescoping series that leaves you with some finite number, which goes to zero as we take the limit as $N$ goes to infinity. The last term goes to zero clearly. Then the middle term goes to $1/2$ clearly.
A: Split the following series into easier to compute series:
$$
\lim_{N\to\infty} \frac{1}{2N+1} \sum_{n=-N}^{N} \left( \frac{1+\cos\left(\frac{\pi}{2}n \right)}{2}\right) =
\lim_{N\to\infty} \frac{1}{2N+1} \left(\sum_{n=-N}^{N} \frac{1}{2} + \sum_{n=-N}^{N} \frac{\cos\left(\frac{\pi}{2}n \right)}{2} \right)=
$$
Since the cosine is periodic every $4n$ the sum $ \left(\cos(0)+\cos(\pi/2)+\cos(\pi)+\cos(-\pi/2)\right)=0$. Then, if even if $N$ were not a multiple of $4$ your remainder will be finite. Some finite term over the infinite $2N+1$ is 0.
You are only left with the first sum. It is the addition of $2N+1$ times $1/2$ so you could cancel the sum with the denominator.
A: Since cosine is an even function we can rewrite the sum as:$$lim_{N\to\infty}\frac{1}{2N+1}2\sum_{n=0}^N\frac{1}{2}+\frac{cos(\frac{\pi n}{2})}{2}$$
Next we can split the sum so that:$$lim_{N\to\infty}\frac{1}{2N+1}[N+\frac{1}{2}\sum_{n=0}^N cos(\frac{\pi n}{2})]$$
Notice that $cos(\frac{\pi n}{2})$ is $0$ at every odd integer. Furthermore, for every pair of even integers $n$, one will result in the cosine term being $1$ and the other being $-1$. So it is a telescoping sum and evaluates to $0$.
Your problem becomes: $$lim_{N\to\infty}\frac{N}{2N+1}=\frac{1}{2}$$
