I am given this series: $\sum_{n=1}^{\infty}\cos(\frac{n\pi}{6})$ and asked if it converges and if it's Cesaro summable or not.
I can easily show that this series diverges. However, I am unsure how to proceed with showing that this is Cesaro summable or not. Here is what I have so far:
Let $\sigma_m=\frac{1}{m}\sum_{n=1}^{m}\cos(\frac{n\pi}{6})$ where $$\{\sigma_m\}=\frac{\cos(\frac{\pi}{6})+\cos(2\frac{\pi}{6})+\cos(3\frac{\pi}{6})+\dotsb+\cos(m\frac{\pi}{6})}{m},$$
and then I have no clue how to proceed. Any help is greatly appreciated!
The sequence $(\cos n\pi /6)_n$ is periodic :$(\sqrt 3 /2,1/2,0,-1/2,-\sqrt 3/2,-1, -\sqrt 3 /2, -1/2,0,1/2, \sqrt 3 /2,1)$ & repeat. And the sum of any $12$ consecutive terms of the sequence is $0.$ So the sequence has bounded partial sums.
BTW there is a general method to get a closed form for $\sum_{j=1}^n \cos jx.$
From $\sin (a+b)-\sin (a-b)=$ $(\sin a\cos b +\sin b \cos a)-(\sin a \cos b-\sin b \cos a)=$ $2\sin b \cos a,$ we have $$2\sin (x/2)\cos jx =\sin (jx+x/2)-\sin (jx-x/2)=\sin (jx+x/2)-\sin ((j-1)x+x/2).$$ So $2\sin (x/2)\sum_{j=1}^n \cos jx=$ $\sum_{j=1}^n(\sin(jx+x/2)-\sin ((j-1)x+x/2).$ This last sum telescopes to $\sin (nx+x/2)-\sin (x/2).$
So if $\sin (x/2)\ne 0$ then $\sum_{j=1}^n\cos jx=\frac {\sin (nx +x/2)}{2\sin (x/2)}-\frac {1}{2}.$
If $\sin (x/2)=0$ then $x/\pi \in \mathbb Z.$