Solutions

(a) Consider a sequence $x_{n}$ that converges to $x$ and another sequence

$$y_{n} = \frac{x_1 + x_2 + \dots + x_{n}}{n}.$$

By definition, for any $\epsilon > 0$,

$$\exists N_1 \in \mathbb{N}, \forall n \in \mathbb{N}, \quad n \geq N_1 \implies \abs{x_{n} - x} < \epsilon.$$

Let's use the $N_1$ from that definition, for any given $\epsilon$. Then, consider the absolute difference

$$\begin{aligned} \abs{y_{n} - x} &= \abs{\frac{x_1 + x_2 + \dots + x_{n}}{n} - x} \\ &= \abs{\frac{x_1 + x_2 + \dots + x_{n}}{n} - \frac{xn}{n}} \\ &= \abs{\frac{(x_1 - x) + (x_2 - x) + \dots + (x_{n} - x)}{n}} \\ &= \abs{ \frac{1}{n}\sum^{N_1 - 1}_{i=1} (x_{i} - x) + \frac{1}{n}\sum^{n}_{i=N_1} (x_{i} - x)} \end{aligned}$$

for $n \geq N_1$. By separating the sum into two parts, we first apply the triangle inequality and then find bounds on both parts, hoping they can be strictly inferior to $\epsilon$ (to satisfy the definition of convergence). Thus,

$$\abs{ \frac{1}{n}\sum^{N_1 - 1}_{i=1} (x_{i} - x) + \frac{1}{n}\sum^{n}_{i=N_1} (x_{i} - x)} \leq \frac{1}{n} \sum^{N_1 - 1}_{i=1} \abs{x_{i} - x} + \frac{1}{n} \sum^{n}_{i=N_1} \abs{x_{i} - x}$$

Notice that the first part has a constant numerator but an increasing denominator, dependent only on $n$. Therefore, by the Archimedean property, we can find $N_2$ for which

$$\frac{1}{n} \sum^{N_1 - 1}_{i=1} \abs{x_{i} - x} < \frac{\epsilon (N_1 - 1)}{N_1} = \frac{\epsilon N_1}{N_1} - \frac{\epsilon}{N_1}$$

when $n \geq N_2$. For the second part, notice that all the terms of the numerator are in $V_{\epsilon}(x)$. Therefore, their sum can be bounded above by $\epsilon$ because the furthest $x_{n}$ can be from $x$ in that interval is less than $\epsilon$. Now, if we choose $N = \max \{N_1, N_2\}$, we obtain

$$\begin{aligned} \abs{y_{n} - x} &=\abs{ \frac{1}{n}\sum^{N_1 - 1}_{i=1} (x_{i} - x) + \frac{1}{n}\sum^{n}_{i=N_1} (x_{i} - x)} \\ &\leq \frac{1}{n} \sum^{N_1 - 1}_{i=1} \abs{x_{i} - x} + \frac{1}{n} \sum^{n}_{i=N_1} \abs{x_{i} - x} \\ &< \frac{\epsilon (N_1 - 1)}{N_1} + \frac{1}{n} (n - N_1 + 1) \epsilon \\ &= \frac{\epsilon N_1}{N_1} - \frac{\epsilon}{N_1} + \epsilon - \frac{\epsilon N}{n} + \frac{\epsilon}{n} \\ &= (\frac{\epsilon N_1}{N_1} -\frac{\epsilon N}{n}) \\ &\leq \epsilon \end{aligned}$$

when $n \geq N$.

(b) Consider the sequence

$$x_{n} = (0, 1, 0, 1, \dots).$$

Clearly, it does not converge. However, the sequence of its average

$$y_{n} = (0, \frac{1}{2}, \frac{1}{3}, \frac{2}{4}, \frac{2}{5}, \frac{3}{6}, \dots)$$

does converge to 1/2.