1
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ Is the numerator for $\sigma_m$ bounded or unbounded? Any special structure? $\endgroup$ – erfink Apr 9 '17 at 1:48
  • $\begingroup$ @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? $\endgroup$ – peachyloaf Apr 9 '17 at 1:58
  • $\begingroup$ 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. $\endgroup$ – erfink Apr 9 '17 at 2:01
  • $\begingroup$ @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. $\endgroup$ – peachyloaf Apr 9 '17 at 2:10
  • $\begingroup$ 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...". $\endgroup$ – erfink Apr 9 '17 at 2:17
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.