To prove that the sequence 1,0,0,1,0,0,1,0,0,... is (C,1) summable:
[A sequence $\{s_n \}$ is (C,1) summable to $L$ if the sequence $\{\sigma_n \}$ converges to $L$ where $$\sigma_n= \frac{s_1+s_2+...s_n}{n}]$$
My attempt: Now for our sequence the general term is $$s_n = \frac{1+2\cos (\frac{2\pi}{3}(n+2))}{3}$$
Now $$\sigma_n= \frac{s_1+s_2+...s_n}{n}$$
I.e. $$\sigma_n= \frac{\sum_{k=1}^{n}s_k}{n}$$
Then $$ \sigma_n = \frac{1}{n} \sum_{k=1}^{n} \frac{1+2\cos (\frac{2\pi}{3}(k+2))}{3}$$
$$\sigma _n= \frac{1}{n} (\sum_{k=1}^{n} \frac{1}{3})+ \frac{2}{3n} \sum_{k=1}^{n} \cos (\frac{2\pi}{3}(k+2))$$
For our sequence i split $\sigma _n $ into 3 cases
Case 1:$ n=1,4,7,12,...$
For $ n=1,4,7,12,...$we have $$ \sigma_n= \frac{1}{3} + \frac{2}{3} = 1$$
Case 2: $n = 2,5,8,...$
Here $$ \sigma_n= \frac{1}{3} + \frac{-1}{3} = 0$$
Similarly case 3: $n = 3,6,9,12,...$
$$ \sigma_n= \frac{1}{3} + \frac{-1}{3} = 0$$
But for $\sigma _n $ to converge the limits must be same in all cases but it is not. Is there any mistake in my calculation or any other way to prove the sequence is (C,1) summable?