Théorèmes de comparaison.
Comparison theorems are among the fundamental theorems for the study of numerical sequences/series:
Theorem: Let \(\sum_{n\in\mathbb{N}} u_n\) and \(\sum_{n\in\mathbb{N}} v_n\) be two series with not negative real terms (the theorem applies more generally to series with terms of constant sign from a certain rank).
i.
Assume: \[\exists n_0\in\mathbb{N}\quad \forall n\geqslant n_0,\quad u_n\leqslant v_n\] Then we have: \[\sum_{n\in\mathbb{N}} v_n \text{ converges } \Rightarrow \sum_{n\in\mathbb{N}} u_n \text{ converges}\]
ii.
Assume: \[u_n\underset{}{=}o(v_n)\] Then we have: \[\sum_{n\in\mathbb{N}} v_n \text{ converges } \Rightarrow \sum_{n\in\mathbb{N}} u_n \text{ converges}\]
iii.
Assume: \[u_n \sim v_n\] Then we have: \[\sum_{n\in\mathbb{N}} v_n \text{ converges } \Leftrightarrow \sum_{n\in\mathbb{N}} u_n \text{ converges}\]
Moreover in the convergence case we have equivalence of remainders: \[\sum_{k=n}^{+\infty}u_k \sim \sum_{k=n}^{+\infty}v_k\] In the divergence case, we have equivalence of partial sums: \[\sum_{k=0}^{n}u_k \sim \sum_{k=0}^{n}v_k\]
Definition - reminders
It is advisable to keep in mind all the formulations for the equivalence of two sequences: \(u_n\sim v_n\) iff \(u_n-v_n=o(v_n)\) iff for every \(\epsilon>0\) (however small), there exists an integer \(n_0\) such that \(\forall n\geqslant n_0\)\[ \begin{alignat*}{1} &\abs{u_n-v_n} \leqslant \epsilon \abs{v_n} \\ &\Leftrightarrow\quad (1-\epsilon) v_n \leqslant u_n \leqslant (1+\epsilon) v_n \qquad(\text{ si }v_n\geqslant0)\\ &\Leftrightarrow\quad 1-\epsilon \leqslant \frac{u_n}{v_n} \leqslant 1+\epsilon \\ &\Leftrightarrow\quad \abs{\frac{u_n}{v_n}-1} \leqslant \epsilon \\ \end{alignat*} \] the last two formulations being valid when \(v_n\) not 0 for \(n\) large, and equivalent to \(\frac{u_n}{v_n}\to 1\).
Of course, we can write the same inequalities by swapping the roles of \(u_n\) and \(v_n\).
Note
Since the properties in question are asymptotic, adapting the proof to the case of \(u_n,v_n\geqslant 0\) from a certain rank \(N\) is straightforward, simply by choosing indices \(n_0,n_1\) greater than \(N\).
The case \(u_n,v_n\leqslant 0\) is handled by considering \(\sum_{n\in\mathbb{N}}-u_n\) and replacing assumption i by \(v_n\leqslant u_n\).
i.
Assume that the series \(\sum_{n\in\mathbb{N}}v_n\) converges. Let \(n\geqslant n_0\). \[ \begin{alignat*}{1} \sum_{k=0}^n u_k &= \sum_{k=0}^{n_0-1} u_k + \sum_{k=n_0}^n u_k \\ &\leqslant \sum_{k=0}^{n_0-1} u_k + \sum_{k=n_0}^n v_k \\ &\leqslant \sum_{k=0}^{n_0-1} u_k + \sum_{k=0}^n v_k \\ &\leqslant \sum_{k=0}^{n_0-1} u_k + \sum_{k=0}^{+\infty} v_k \\ \end{alignat*} \] The upperbound remains valid for \(n < n_0\).
This shows that the sequence of partial sums \((S_n)\) is bounded. As \(u_n\geqslant0\), the sequence \((S_n)\) is also increasing, so it converges, i.e. the series \(\sum_{n\in\mathbb{N}}u_n\) converges.
ii.
Let \(\epsilon=1\). Since \(u_n\underset{}{=}o(v_n)\), \[\exists n_0\in\mathbb{N}\quad \forall n\geqslant n_0,\quad u_n=\abs{u_n}\leqslant \epsilon\abs{v_n}=v_n\] which brings us back to the previous case.
iii.
Assume that the series \(\sum_{n\in\mathbb{N}}v_n\) converges.
Let \(\epsilon>0\). \[\exists n_0\in\mathbb{N}\quad\forall n\geqslant n_0,\quad \abs{u_n-v_n}\leqslant \epsilon v_n\] Thus, \(\forall n\geqslant n_0\), \[ \begin{alignat*}{1} \sum_{k=0}^n u_k &= \sum_{k=0}^{n_0-1} u_k + \sum_{k=n_0}^n u_k \\ &\leqslant \sum_{k=0}^{n_0-1} u_k + (1+\epsilon) \sum_{k=n_0}^n v_k\\ &\leqslant \sum_{k=0}^{n_0-1} u_k + (1+\epsilon) \sum_{k=0}^{+\infty} v_k \\ \end{alignat*} \] This remains valid for \(n < n_0\).
This shows that the sequence of partial sums \((S_n)\) is bounded from above. As \(u_n\geqslant0\), the sequence \((S_n)\) is also increasing, so it converges, i.e. the series \(\sum_{n\in\mathbb{N}}u_n\) converges.
By symmetry, we also have \[\sum_{n\in\mathbb{N}}u_n\text{ converges }\Rightarrow \sum_{n\in\mathbb{N}}v_n \text{ converges }\]
Suppose the series diverge. In this case the sequences of partial sums are not bounded and \[\lim_{n\to +\infty} S_n =\lim_{n\to +\infty} T_n = + \infty\]
When \(n\geqslant n_0\), \[ \begin{alignat*}{1} \sum_{k=0}^{n_0-1} u_k + (1-\epsilon) \sum_{k=n_0}^n v_k &\leqslant \sum_{k=0}^n u_k \leqslant \sum_{k=0}^{n_0-1} u_k + (1+\epsilon) \sum_{k=n_0}^n v_k\\ \Leftrightarrow\quad \underbrace{\sum_{k=0}^{n_0-1} u_k -(1-\epsilon) \sum_{k=0}^{n_0-1} v_k}_{=A} + (1-\epsilon) \sum_{k=0}^n v_k&\leqslant \sum_{k=0}^n u_k \leqslant \underbrace{\sum_{k=0}^{n_0-1} u_k - (1+\epsilon) \sum_{k=0}^{n_0-1} v_k}_{=B} + (1+\epsilon) \sum_{k=0}^n v_k\\ \end{alignat*} \]
The expressions \(A\) and \(B\) are constant (do not depend on \(n\)), so they are negligible compared with \(T_n\). We can therefore find an integer \(n_1\geqslant n_0\) such that \(\forall n\geqslant n_1\),\[ \begin{alignat*}{1} -\epsilon \sum_{k=0}^n v_k + (1-\epsilon) \sum_{k=0}^n v_k &\leqslant \sum_{k=0}^n u_k \leqslant \epsilon \sum_{k=0}^n v_k + (1+\epsilon) \sum_{k=0}^n v_k\\ \Leftrightarrow\quad (1-2\epsilon) \sum_{k=0}^n v_k &\leqslant \sum_{k=0}^n u_k \leqslant (1+2\epsilon) \sum_{k=0}^n v_k\\ \end{alignat*} \] Which proves \(S_n\sim T_n\).
If we now assume that the series converge, the previous reasoning is no longer valid, but for \(n\geqslant n_0\), by summing the relations between \(u_n\) and \(v_n\) up to infinity, we can write \[\begin{alignat*}{1} (1-\epsilon) \sum_{k=n}^{+\infty} v_k &\leqslant \sum_{k=n}^{+\infty} u_k \leqslant (1+\epsilon) \sum_{k=n}^{+\infty} v_k\ \end{alignat*} \] which proves that the remainders of the series are equivalent.
A numerical series is a sequence, the sequence of partial sums \((S_n)_{n\in\mathbb{N}}\).
Conversely, we tend to think less of it, but any sequence can always be written as a series: \[ u_n = \sum_{k=1}^n (u_k-u_{k-1}) + u_0 \] This series has the same nature as the sequence \((u_n)\). If \(u_n\to l\), we can also write the sequence as a remainder of a convergent series: \[ -u_n = \sum_{k=n}^{+\infty} (u_{k+1} - u_k) -l \]
In this way, we can often apply the partial sum/remainder equivalence theorem to a sequence when we're looking for an equivalent:
- If \(u_n\to l\), we'll try to apply the equivalence of remainders.
- If \(u_n \to \infty \), we'll try to apply the equivalence of partial sums
