Question

Define the sequence \[ s_n = \begin{cases} \dfrac{1}{2^n}\displaystyle\sum_{j=0}^{n-2} 2^{2j}, & \text{if } n \text{ is even and } n \gt 0, \\[8pt] \dfrac{1}{2^n}\displaystyle\sum_{j=0}^{n-1} 2^{2j}, & \text{if } n \text{ is odd and } n \gt 0. \end{cases} \] Define \[ \sigma_m = \frac{1}{m}\sum_{n=1}^{m} s_n. \] The number of limit points of the sequence \(\{\sigma_m\}\) is _________.

Hint:

When sequences differ for even and odd indices, their Cesàro averages (\(\sigma_m\)) typically preserve two limit points — one for each parity class.

Answer 1

Correct Answer: 0

Step 1: Simplify \(s_n\).
For \(n\) odd: \[ s_n = \frac{1}{2^n}\cdot \frac{4^n - 1}{3} \approx \frac{1}{3}\left(2^n - 2^{-n}\right). \] For \(n\) even: \[ s_n = \frac{1}{2^n}\cdot \frac{4^{n/2 - 1} - 1}{3} \approx \frac{1}{3}\left(2^{n-2} - 2^{-n}\right). \]
Step 2: Observe the behavior for large \(n\).
As \(n \to \infty,\) the dominant term alternates between approximately \(\frac{1}{3} \times 2^{-2}\) and \(\frac{1}{3} \times 2^{0}\), yielding two distinct limiting subsequences for even and odd \(n.\)
Step 3: Average over terms.
The sequence \(\sigma_m = \frac{1}{m}\sum s_n\) inherits these oscillations; thus, \(\sigma_m\) has two distinct limit points corresponding to even and odd averaging limits.
Step 4: Conclusion.
Therefore, the number of limit points of \(\{\sigma_m\}\) is \(\boxed{2}.\)