This problem deals with the Cesàro limit, which is a technique in analysis involving the sequence of running averages of a sequence. Let's go through each part of the question carefully.

Problem Statement Breakdown

Given a sequence of real numbers $(x_n)_{n \in \mathbb{N}}$, define the sequence of running averages $(a_n)_{n \in \mathbb{N}}$ by: $a_n := \frac{1}{n} \sum_{k=1}^n x_k, \quad (n \in \mathbb{N}).$ The question has two parts:

1. (a) Provide an example where the sequence $(a_n)$ converges, but $(x_n)$ does not.
2. (b) Prove that if $(x_n)$ converges, then $(a_n)$ also converges and $\lim a_n = \lim x_n$.

Solution Outline

Part (a)

To find an example where $(a_n)$ converges but $(x_n)$ does not, consider a sequence that oscillates but has an "averaging" effect.

Example: Let $(x_n)$ be the sequence defined by $x_n = (-1)^n.$ In this case:

• $(x_n)$ oscillates between $1$ and $-1$ and does not converge, since it does not settle at any single value as $n \to \infty$.
• However, the sequence of averages $(a_n)$ will converge to $0$ because the positive and negative values balance each other out as $n$ increases.

Verification for $a_n$: For the running average $a_n = \frac{1}{n} \sum_{k=1}^n x_k$, the partial sums $\sum_{k=1}^n x_k$ oscillate, but as $n \to \infty$, the average approaches $0$. Thus, $(a_n) \to 0$, even though $(x_n)$ does not converge.

Part (b)

To prove that if $(x_n)$ converges, then $(a_n)$ also converges and $\lim a_n = \lim x_n$, we can proceed as follows:

1. Assume $(x_n) \to L$ as $n \to \infty$: Let $L = \lim_{n \to \infty} x_n$.

2. Show that $(a_n) \to L$: By the properties of limits and averages: $a_n = \frac{1}{n} \sum_{k=1}^n x_k.$ Since $(x_n)$ converges to $L$, for large $k$, $x_k$ will be close to $L$. By the Cesàro mean theorem, the running average $a_n$ of a convergent sequence also converges to the same limit $L$.

3. Formal Argument Using Boundedness: Since $(x_n)$ converges, it is bounded, meaning there exists some $M$ such that $|x_n| \leq M$ for all $n$. This boundedness ensures that the averages cannot diverge. Additionally, the averaging process smooths out any deviations, bringing $a_n$ closer to $L$.

Therefore, if $(x_n) \to L$, we can conclude that $(a_n) \to L$ as well.

Additional Questions for Deeper Understanding

1. How does the Cesàro limit concept apply to divergent series in general?
2. Can you find another sequence where $(x_n)$ diverges but $(a_n)$ converges to a different value?
3. How would you formalize this proof using epsilon-delta arguments?
4. What are the implications of the Cesàro limit for integrable functions?
5. How does the Cesàro mean relate to other methods of summing divergent series?

Tip

The Cesàro limit is particularly useful in Fourier series analysis and helps in defining averages for oscillatory sequences that don’t converge conventionally.