Cesàro Theorem

Let $(a_n)_{n \in \mathbb{N}^*}$ be a sequence of complex numbers such that $\lim_{n \to \infty} a_n = \ell \in \mathbb{C}$. Then:

$$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n a_k = \ell$$

Setting $\forall n \in \N^{*}, b_n = a_n - \ell$, we can assume without loss of generality that $\ell = 0$.

Let $\varepsilon > 0$. Since $b_n \to 0$, there exists $N \in \mathbb{N}^*$ such that:

$$\forall k > N, \quad |b_k| \le \frac{\varepsilon}{2}$$

For any $n > N$, split the sum into two parts:

$$\left| \frac{1}{n} \sum_{k=1}^n b_k \right| \le \frac{1}{n} \sum_{k=1}^N |b_k| + \frac{1}{n} \sum_{k=N+1}^n |b_k| \le \frac{M_N}{n} + \frac{n - N}{n} \frac{\varepsilon}{2} < \frac{M_N}{n} + \frac{\varepsilon}{2}$$

where $M_N = \sum_{k=1}^N |b_k|$ is a fixed constant independent of $n$.

Since $\lim_{n \to \infty} \frac{M_N}{n} = 0$, there exists $N' \ge N$ such that:

$$\forall n > N', \quad \frac{M_N}{n} < \frac{\varepsilon}{2}$$

Thus, for all $n > N'$:

$$\left| \frac{1}{n} \sum_{k=1}^n a_k - \ell \right| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$$

For the second question, same reasoning.

As counterexamples, you can consider $u_{n}=(-1)^{n}$ and $f=sin$.

Still, there are some converse results such as when the sequence / function is real and monotonous (or more generally if we already know that it has a limit - finite or not).