# Show a series is Cesaro summable.

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!

• Is the numerator for $\sigma_m$ bounded or unbounded? Any special structure? – erfink Apr 9 '17 at 1:48
• @erfink It's bounded below by -1 and above by 1. So the sum of the numerator is less than $m$. In fact, the sum is also bounded due to the nature of the cosine function. Would that mean the partial sequence $\{\sigma_m\}$ converge to 0? – peachyloaf Apr 9 '17 at 1:58
• Close. It is bounded, but you'll want to be a little more careful about the precise bounds. Once you find those bounds, it's all squeeze theorem. – erfink Apr 9 '17 at 2:01
• @erfink Would the precise bounds be $-\frac{3}{2}-\frac{\sqrt{3}}{2}$ and $\frac{1}{2}+\frac{\sqrt{3}}{2}$? Would I apply the squeeze theorem such as: Since the numerator is bounded but $m$ is unbounded, we have the partial sequence converging to zero, hence it's Cesaro summable? Sorry if I seem a bit clueless. – peachyloaf Apr 9 '17 at 2:10
• Yep, those are the bounds that I get. Then we have $\forall m$ that $\frac{1}{m} (\frac{-3-\sqrt{3}}{2}) \leq \sigma_m \leq \frac{1}{m}(\frac{1 + \sqrt{3}}{2})$ and taking a limit as $m \to \infty$ gives the desired result. Also, "sequence of partial sums converging to..." instead of "partial sequence converging to...". – erfink Apr 9 '17 at 2:17

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 $x/\pi \in \mathbb Z.$