Number of limit points of Abel sum of given sequence? Define the sequence
$$
s_{n}= \begin{cases}\frac{1}{2^{n}} \sum_{j=0}^{n-2} 2^{2 j} & \text { if } n>0 \text { is even, } \\ \frac{1}{2^{n}} \sum_{j=0}^{n-1} 2^{2 j} & \text { if } n>0 \text { is odd }\end{cases}
$$
Define $\sigma_{m}=\frac{1}{m} \sum^{m} s_{n}$. Then how to calculate number of limit points of the sequence $\left\{\sigma_{m}\right\}$?
By seeing this problem, it looks to complicated. Is there any way to find number of limit point? Even if some one gives me any hint that would be very nice.
Thank you so much
 A: I assume that you are allowed to use the fact that each limit point of a sequence is a limit of a subsequence and that it should be $\sigma_{n}=\frac{1}{n} \sum^{n} s_{n}$.

Let's consider a subsequence $\left(\sigma_{n_i}\right)_{i\in\mathbb{N}}$, where $n_i=2i$. Then we use a telescope sum:
$$\sigma_{n_i}= \frac{1}{n_i}\frac{1}{2^{n_i}}\sum\limits_{j=0}^{n_i-2} 2^{2 j}= \frac{1}{n_i}\frac{1}{2^{n_i}}\frac{1}{1-2^2}\left(2^0-2^{2n_i-2}\right)= \frac{1}{n_i}\frac{1}{3\cdot2^{n_i}}\left(2^{2n_i-2}-1\right)=\frac{1}{n_i}\frac{1}{3}\left(2^{n_i-2}-\frac{1}{2^{n_i}}\right).$$
We can immediately see that $\left(\sigma_{n_i}\right)_{i\in\mathbb{N}}$ is unbounded.
The same argument can be applied to the case where $n_i=2i+1$.
Any other subsequence contains infinitely many members of one or both of the above mentioned subsequences (because  they are a bijection into the set of all members of the original sequence $\sigma_n$). Hence, all subsequences are unbounded and we conclude that there don't exist any limit points.
