This problem is about the Cesàro limit and deals with the concept of convergence of sequences and their averages. Let's address each part of the question in detail.

Part (a): Example where $(a_n)$ converges but $(x_n)$ does not

We need to find a sequence $(x_n)$ that does not converge, but whose sequence of averages $(a_n)$ defined by $a_n = \frac{1}{n} \sum_{k=1}^n x_k$ does converge.

A classic example is the sequence $x_n = (-1)^n$, which oscillates between 1 and -1:

• $x_1 = 1$
• $x_2 = -1$
• $x_3 = 1$
• $x_4 = -1$
• and so on.

This sequence does not converge because it keeps oscillating and does not settle to a single value. However, let's examine the sequence of averages $(a_n)$: $a_n = \frac{1}{n} \sum_{k=1}^n x_k.$

For $x_n = (-1)^n$:

• When $n$ is even, the sum $\sum_{k=1}^n x_k$ is 0, and hence $a_n = 0$.
• When $n$ is odd, the average $a_n$ gets closer and closer to 0 as $n$ increases.

Thus, $a_n \to 0$ as $n \to \infty$, even though $(x_n)$ does not converge. This provides the required example.

Part (b): Prove that if $(x_n)$ converges, then $(a_n)$ also converges, and $\lim a_n = \lim x_n$.

Assume that $(x_n)$ converges to some limit $L$, i.e., $\lim_{n \to \infty} x_n = L.$

Since $(x_n)$ converges, it is also bounded (as the hint suggests). Let’s examine the behavior of the averages $a_n$: $a_n = \frac{1}{n} \sum_{k=1}^n x_k.$

To show that $a_n$ converges to $L$, observe that for large $n$, most terms $x_k$ in the sum $\sum_{k=1}^n x_k$ will be close to $L$ because $x_k \to L$ as $k \to \infty$. Intuitively, the average of values that are close to $L$ should also be close to $L$.

To make this rigorous, we can use the Cesàro Mean Theorem, which states that if $x_n \to L$, then $a_n \to L$ as well. Here’s a brief outline of the proof:

1. Splitting the Sum: Write $x_k = L + (x_k - L)$, so $a_n = \frac{1}{n} \sum_{k=1}^n x_k = \frac{1}{n} \sum_{k=1}^n L + \frac{1}{n} \sum_{k=1}^n (x_k - L).$ The first term simplifies to $L$ because $\frac{1}{n} \sum_{k=1}^n L = L.$

2. Remainder Term: For the second term, we have $\frac{1}{n} \sum_{k=1}^n (x_k - L).$ Since $x_k \to L$, for any $\epsilon > 0$, there exists $N$ such that for $k \ge N$, $|x_k - L| < \epsilon$. Thus, as $n \to \infty$, this remainder term goes to 0.

3. Conclusion: Therefore, $a_n \to L$ as $n \to \infty$, which proves that if $x_n \to L$, then $a_n \to L$ as well.

This completes the proof that if $(x_n)$ converges, then $(a_n)$ also converges, with $\lim a_n = \lim x_n$.

Would you like further details on any part of this solution?

Here are some related questions for deeper understanding:

1. What is the significance of boundedness in proving convergence for the Cesàro limit?
2. How does the Cesàro Mean Theorem apply to non-convergent sequences?
3. Can the Cesàro limit be applied to sequences that diverge to infinity?
4. How does the averaging process affect sequences with large oscillations?
5. Can the concept of the Cesàro limit be extended to double sequences or series?

Tip: When analyzing convergence of sequences with oscillating behavior, consider averaging methods like Cesàro sums to simplify convergence analysis.