Does the Series $\sum_{n=1}^{\infty} (1-\cos\frac{\pi}{n})$ Converge? 
Does the serie $\displaystyle\sum_{n=1}^{\infty}\Bigl(1-\cos\frac{\pi}{n}\Bigr)$ converge?

Limit test of $1-\cos\frac{\pi}{n}$
$$
\lim_{x\to\infty} 1-\cos\frac{\pi}{n} =
1 -\lim_{n\to\infty}\cos\frac{\pi}{n}= 1-1 = 0
$$
I've checked the necessary condition that $\lim_{n\to\infty} 1-\cos\frac{\pi}{n} = 0 $
But how do I check if $\sum_{n=1}^{\infty}(1-\cos\frac{\pi}{n})$ converges?
 A: Hint:
$$
0<1-\cos\frac{\pi}{n}=2\sin^2\frac{\pi}{2n}
$$
Now recall that, for $x>0$, $\sin x<x$.

And no, your attempt to show divergence is not good.
A: Note that for n large the sum from n to infinity is bounded above by the result of substituting n with x and integrating the same function f(x) with respect to x from n to infinity, viewing the sum as a Riemann sum. Substitute $x = 1/u$, $dx = -du/u^2$, to get that this integral is equal to that of $(1 - cos(u))/u^2$ from 0 to 1/n. You can use the Taylor series for cos(u) to show that the integrand approaches 1/2 as u goes to zero. The integrand is continuous on (0, 1/n), so as n goes to infinity the integral goes to 1/2n. This implies that the tail of the sum goes to zero as n grows, and since the sum from 1 to n is finite the sum converges.
A: Hint: $\cos(x) = 1 + O(x^2)$ for $x \to 0$ by Taylor's theorem.
A: You can use L'hopital's rule (twice) to show that
$$\lim_{x \rightarrow 0} {1 - \cos x \over x^2} = {1 \over 2}$$
So you can use the limit comparison test with the series ${\displaystyle \sum_{n = 1}^{\infty} {1 \over n^2}}$ to show your series converges.
