Let \( (u_n)_{n \in \mathbb{N}} \) be a numerical sequence.
Let use suppose that \((u_n)\) tends towards a certain limit \(\Bigl(l \in \bigl\{ \mathbb{R} \cup \infty \bigr \}\Bigr)\) reel or infinite, from a certain rank \(n_0\):
$$ \forall \varepsilon > 0, \ \exists n_0 \in \mathbb{N}, \ \forall n \geqslant n_0, \hspace{2em} | u_n - l | < \frac{\varepsilon}{2} < \varepsilon \qquad (H) $$
For the sake of rigor in the demonstration, we need to find a higher bound by \(\frac{\varepsilon}{2}\), because later in the demonstration we will need to major two parts.
Now, if we look at how its average behaves:
$$ \left| \frac{1}{n} \sum_{k = 1}^n u_k - l \right| = \left| \frac{u_1 + u_2 + \ ... \ + \hspace{0.2em} u_n}{n} - l \right| $$
Putting them over a common denominator, we do have:
$$ \left| \frac{1}{n} \sum_{k = 1}^n u_k - l \right| = \left| \frac{u_1 + u_2 + \ ... \ + \hspace{0.2em} u_n}{n} - \textcolor{rgb(118 139 240)}{\frac{n}{n} \times \ } l \right| $$
$$ \left| \frac{1}{n} \sum_{k = 1}^n u_k - l \right| = \left| \frac{u_1 + u_2 + \ ... \ + \hspace{0.2em} u_n - nl}{n} \right| $$
$$ \left| \frac{1}{n} \sum_{k = 1}^n u_k - l \right| = \left| \frac{(u_1 - l) + (u_2 - l) \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} (u_n - l)}{n} \right| $$
But, we know that:
$$ \forall (a,b) \in \mathbb{R}^2, $$
$$ | a + b | \leqslant |a| + |b| $$
So,
$$ \left| \frac{1}{n} \sum_{k = 1}^n u_k - l \right| \leqslant \frac{|u_1 - l| + |u_2 - l| \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} |u_n - l|}{n} $$
Let us divide the right member in two distinct parts:
$$ \left| \frac{1}{n} \sum_{k = 1}^n u_k - l \right| \leqslant \frac{1}{n} \sum_{k = 1}^{n_0 - 1} |u_k - l| + \frac{1}{n} \sum_{k = n_0}^n |u_k - l| $$
However, we saw with \((H)\) that from a certain rank \(n_0\), the sequences \((u_n)\) converges.
$$ \left| \frac{1}{n} \sum_{k = 1}^n u_k - l \right| \leqslant \frac{|u_1 - l| + |u_2 - l| + \ ... \ + \hspace{0.2em} |u_{n_0 - 1} - l| \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} \frac{\varepsilon}{2} + \frac{\varepsilon}{2} + \frac{\varepsilon}{2} \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} \frac{\varepsilon}{2}}{n} $$
For simplicity, we define the following variable \(M\), sum of the preceding terms \(u_{n_0}\) :
$$ M = |u_1 - l| + |u_2 - l| + \ ... \ + \hspace{0.2em} |u_{n_0 - 1} - l| $$
which we substitute in the previous expression.
$$ \left| \frac{1}{n} \sum_{k = 1}^n u_k - l \right| \leqslant \frac{M}{n} + \frac{(n - n_0 + 1)\varepsilon}{2n} $$
At this stage, we must choose a rank \(n_1\) large enough to ensure that the first block becomes strictly less than \(\frac{\varepsilon}{2}\) (the right-hand side being already under direct control thanks to hypothesis \((H)\)).
Since \(M\) is a constant independent of \(n\), the first block tends toward \(0\) as \(n\) approaches infinity:
$$ \lim_{n \to \infty} \frac{M}{n} = 0 $$
Consequently, by the definition of a limit, there exists a rank \(n_1\) from which this block becomes strictly less than \(\frac{\varepsilon}{2}\):
$$ \exists n_1 \in \mathbb{N}, \ \forall n \geqslant n_1, \hspace{2em} \frac{M}{n} < \frac{\varepsilon}{2} $$
Let us set:
$$ N = max\Bigl \{ n_0, \ n_1\Bigr \}$$
Thus, by summing both bounds:
$$ \forall (n \geqslant N), $$
$$ \left| \frac{1}{n} \sum_{k = 1}^n u_k - l \right| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} $$
$$ \left| \frac{1}{n} \sum_{k = 1}^n u_k - l \right| < \varepsilon $$
$$ \exists n_1 \in \mathbb{N}, \ \forall n \geqslant n_1, $$
Let us set:
$$ N = max\Bigl \{ n_0, \ n_1\Bigr \}$$
Thus, by summing both bounds:
$$ \forall (n \geqslant N), $$
$$ \left| \frac{1}{n} \sum_{k = 1}^n u_k - l \right| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} $$
$$ \left| \frac{1}{n} \sum_{k = 1}^n u_k - l \right| < \varepsilon $$
Then, if \((u_n)\) tends towards a certain limit, its average tends towards that same limit.
$$ \lim_{n \to \infty} \bigl[ u_n \bigr] = l \Longrightarrow \lim_{n \to \infty} \left[ \frac{1}{n} \sum_{k = 1}^n u_k \right] = l \qquad \bigl(\text{Cesàro's theorem} \bigr) $$
Let \( (u_n)_{n \in \mathbb{N}} \) be a non-zero recurrent numerical sequence of order 2, and linear combination of its preceding terms such as:
$$ u_{n + 2} = p.u_{n + 1} + q.u_n \qquad (1) $$
$$ \Bigl(\text{ avec } (p, q) \in \bigl(\mathbb{R}^*\bigr)^2 \Bigr) $$
For this type of recursive sequence, exponents work very well with exponential solutions. Therefore, a suitable solution would be of the form:
$$ u_n = r^n \qquad (\text{ avec } r \in \mathbb{R}^*) \qquad (2) $$
So, if we inject \((2)\) into \((1)\), we do have this:
$$ r^{n + 2} = p.r^{n + 1} + q.r^n $$
Dividing by \(r^n\), we now obtain that:
$$ r^2 = pr + q $$
$$ r^2 - pr - q = 0 \qquad(E_c) $$
At this stage, we are left with a
quadratic equation
to solve.
We then obtain at most two solutions for \(r\): \((\alpha, \beta)\).
However, these two solutions sought, \(\alpha^n\) and \(\beta^n\) form a two-dimensional
vector space
on \(\mathbb{R}\).
Because, if both \(\alpha^n\) and \(\beta^n\) are solutions for being \((u_n)\), then:
$$ \left \{ \begin{gather*} \alpha^{n + 2} = p.\alpha^{n + 1} + q.\alpha \\ \\ \beta^{n + 2} = p.\beta^{n + 1} + q.\beta \end{gather*} \right \} $$
$$ \left \{ \begin{gather*} A\alpha^{n + 2} = Ap.\alpha^{n + 1} + Aq.\alpha \qquad(3) \\ \\ B\beta^{n + 2} = Bp.\beta^{n + 1} + Bq.\beta \qquad(4) \end{gather*} \right \} $$
Performing the operation \( \bigl((3) + (4) \bigr)\), we get that:
$$ A\alpha^{n + 2} + B\beta^{n + 2} = Ap.\alpha^{n + 1} + Aq.\alpha + Bp.\beta^{n + 1} + Bq.\beta $$
Now, gathering terms according to their degree,
$$ A\alpha^{n + 2} + B\beta^{n + 2} = Ap.\alpha^{n + 1} + Bp.\beta^{n + 1} + Aq.\alpha + Bq.\beta $$
$$ \underbrace{A\alpha^{n + 2} + B\beta^{n + 2}} _{u_{n + 2}} = p\underbrace{(A.\alpha^{n + 1} + B.\beta^{n + 1})} _{u_{n + 1}} + q\underbrace{(A.\alpha + B.\beta)} _{u_n} $$
We clearly see that the following sequence:
$$ u_n = A.\alpha^n + B.\beta^n $$
is definitely solution \((1)\).
Thus, we cand define the general term of this sequence:
$$ u_{n + 2} = p.u_{n + 1} + q.u_n \Longrightarrow u_n = A. \alpha^n + B.\beta^n \hspace{2em} \text{with } \left \{ \begin{gather*} (p, q) \in \bigl(\mathbb{R}^*\bigr)^2 \\ \\ A \text{ and } B \text{ to determine according to } u_0 \text{ and } u_1 \\ \\ \alpha \text{ and } \beta \text{ the two roots of } \Bigl[ r^2 - pr - q = 0 \Bigr] \end{gather*} \right \} $$
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