The properties of numerical sequences
Let \( (u_n)_{n \in \mathbb{N}} \) be a numerical sequence.
Then, if \((u_n)\) tends towards a certain limit, its average tends towards that same limit.
$$ \lim_{n \to \infty} \bigl[ u_n \bigr] = l \Longrightarrow \lim_{n \to \infty} \left[ \frac{1}{n} \sum_{k = 0}^n u_n \right] = l \qquad \bigl(\text{Cesàro's theorem} \bigr) $$
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:
$$ u_{n + 2} = p.u_{n + 1} + q.u_n \qquad (1) $$
$$ \Bigl(\text{ avec } (p, q) \in \bigl[\mathbb{R}^*\bigr]^2 \Bigr) $$
Then, we cand define the general term of this sequence:
$$ u_{n + 2} = p.u_{n + 1} + q.u_n \Longrightarrow u_n = A. \alpha^n + B.\beta^n
\hspace{2em} \text{with } \left \{ \begin{gather*}
(p, q) \in \bigl[\mathbb{R}^*\bigr]^2 \\ \\
A \text{ and } B \text{ to determine according to } u_0 \text{ and } u_1 \\ \\
\alpha \text{ and } \beta \text{ the two roots of } \Bigl[ r^2 - pr - q = 0 \Bigr]
\end{gather*} \right \}
$$
Demonstrations
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.
So, from a certain rank \(n_0\):
$$ \exists (n_0 \geqslant n), \ \forall(\varepsilon > 0), \hspace{2em} | u_n - l | < \varepsilon < \frac{\varepsilon}{2} \qquad (1) $$
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:
$$ \left| \frac{1}{n} \sum_{k = 0}^n u_n - l \right| = \left| \frac{u_0 + u_1 + u_2 \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} u_n}{n} - l \right| $$
Putting them over a common denominator, we do have:
$$ \left| \frac{1}{n} \sum_{k = 0}^n u_n - l \right| = \left| \frac{u_0 + u_1 + u_2 \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} u_n}{n} - \textcolor{#6F79AB}{\frac{n}{n}}l \right| $$
$$ \left| \frac{1}{n} \sum_{k = 0}^n u_n - l \right| = \left| \frac{u_0 + u_1 + u_2 \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} u_n - nl}{n} \right| $$
$$ \left| \frac{1}{n} \sum_{k = 0}^n u_n - l \right| = \left| \frac{(u_0 - l) + (u_1 - l) + (u_2 - l) \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} (u_n - l)}{n} \right| $$
But, we know that:
$$ \forall (a,b) \in \mathbb{R}^2, $$
$$ | a + b | \leqslant |a| + |b| $$
So,
$$ \left| \frac{1}{n} \sum_{k = 0}^n u_n - l \right| \leqslant \frac{|u_0 - l| + |u_1 - l| + |u_2 - l| \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} |u_n - l|}{n} $$
However, we saw with \((1)\) that from a certain rank \(n_0\), the sequences \((u_n)\) converges.
$$ \left| \frac{1}{n} \sum_{k = 0}^n u_n - l \right| \leqslant \frac{|u_0 - l| + |u_1 - l| + |u_2 - l| \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} |u_{n_0} - l| + |u_{n_0 + 1} - l| + |u_{n_0 + 2} - l| \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} |u_n - l|}{n} $$
$$ \left| \frac{1}{n} \sum_{k = 0}^n u_n - l \right| \leqslant \frac{|u_0 - l| + |u_1 - l| + |u_2 - l| \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} \frac{\varepsilon}{2} + \frac{\varepsilon}{2} + \frac{\varepsilon}{2} \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} \frac{\varepsilon}{2}}{n} $$
For simplicity, we define the following variable \(M\), sum of the preceding terms \(u_{n_0}\) :
$$ M = |u_0 - l| + |u_1 - l| + |u_2 - l| \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} |u_{n_0 - 1} - l| $$
which we substitute in the previous expression.
$$ \left| \frac{1}{n} \sum_{k = 0}^n u_n - l \right| \leqslant \frac{M}{n} + \frac{(n - n_0)\varepsilon}{2n} $$
At this stage, we need to choose a \(n_1\) (depending on \(M\) and then also on \(\varepsilon\)) high enough to ensure that \(\frac{M}{n} < \frac{\varepsilon}{2}\).
The right part is already under control because \(n_0\) is sufficient for this case.
Let us set down:
$$ N = max\Bigl \{ n_0, \ n_1\Bigr \}$$
Hence,
$$ \forall (n \geqslant N), $$
$$ \left| \frac{1}{n} \sum_{k = 0}^n u_n - l \right| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} $$
$$ \left| \frac{1}{n} \sum_{k = 0}^n u_n - l \right| < \varepsilon $$
Then, if \((u_n)\) tends towards a certain limit, its average tends towards that same limit.
$$ \lim_{n \to \infty} \bigl[ u_n \bigr] = l \Longrightarrow \lim_{n \to \infty} \left[ \frac{1}{n} \sum_{k = 0}^n u_n \right] = l \qquad \bigl(\text{Cesàro's theorem} \bigr) $$
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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