Question

Let \( a_k = (-1)^{k-1} \), \( s_n = a_1 + a_2 + \cdots + a_n \), and \( \sigma_n = \frac{1}{n} (s_1 + s_2 + \cdots + s_n) \), where \( k, n \in \mathbb{N} \).

\[ \text{ Then} \,\, \lim_{n \to \infty} \sigma_n = \text{.................} \text{ (correct up to one decimal place)}. \]

Hint:

For alternating series, the partial sums tend to oscillate around a value. The average of these sums tends to the limit of the series.

Answer 1

Correct Answer: 0.4 - 0.6

First, calculate the sequence \( a_k = (-1)^{k-1} \). It alternates: \( a_1 = 1 \), \( a_2 = -1 \), \( a_3 = 1 \), \( a_4 = -1 \), etc. Then, find the partial sums \( s_n = a_1 + a_2 + \cdots + a_n \). This sequence progresses: \( s_1 = 1 \), \( s_2 = 0 \), \( s_3 = 1 \), \( s_4 = 0 \), indicating even \( n \), \( s_n = 0 \) and odd \( n \), \( s_n = 1 \).

Next, compute \( \sigma_n = \frac{1}{n} (s_1 + s_2 + \cdots + s_n) \).

Examine the pattern: for large \( n \), about half \( s_k \)'s are zeros and half are ones (when \( n \) is even), and slightly more ones than zeros when \( n \) is odd. Given this periodic structure, \(\sigma_n\) converges towards:\[ \lim_{n \to \infty} \frac{1}{n} \left(\frac{n}{2}\right) = \frac{1}{2} \].

The correct limit of \( \sigma_n \) is \(\boxed{0.5}\)

Thus, \(\lim_{n \to \infty} \sigma_n = 0.5\), confirming its precision to one decimal place.