Ordering and convergence.

Proposition 3.8.1   Let $ (u_n)$ be a sequence whose terms are all non negative. If this sequence is convergent, with limit $ U$ , then $ U \geq 0$ .

Example 3.8.2   Let $ u_n= \frac 1n, \; n \geq 1$ . Then $ \forall n \geq 1, \; u_n >0$ . We have $ \underset{n \rightarrow \infty}{\text{lim}}u_n=0$ .

Corollary 3.8.3   Let $ (u_n)$ and $ (v_n)$ be two sequences such that for any natural $ n$ , $ u_n < v_n$ (or $ u_n \leq v_n$ ). If both sequences are convergent, with respective limits $ U$ and $ V$ , then $ U \leq V$ .

Example 3.8.4   Let $ u_n= \frac {1}{n^2}, \; n \geq 1$ and $ v_n= \frac 1n, \; n \geq 1$ . Then $ \forall n \geq 2, \; u_n < v_n$ . We have $ \underset{n \rightarrow \infty}{\text{lim}}u_n= \underset{n \rightarrow \infty}{\text{lim}}v_n0$ .

Proposition 3.8.5 (Sandwich theorem)   Let $ (u_n)$ ,$ v_n$ and $ w_n$ be three sequences whose terms verify the condition: $ \forall n, \; u_n < v_n < w_n$ . If this sequences $ (u_n)$ and $ (w_n)$ are convergent, with the same limit $ l$ , then $ (v_n)$ is convergent and its limit is equal to $ l$ .

Example 3.8.6   Let $ a_n=\frac 1n \sin (n! \sqrt{n^2+3})$ . Then for any natural $ n$ , we have: $ -\frac 1n \leq a_n \leq \frac 1n$ . As $ \underset{n \rightarrow \infty}{\text{lim}} \frac 1n = 0$ , we have $ \underset{n \rightarrow \infty}{\text{lim}} a_n =0$ .

Corollary 3.8.7   If the sequence $ (u_n)$ converges towards the limit $ l$ , then the sequence $ (\vert u_n\vert)$ converges towards the limit $ \vert l\vert$ .

The converse of Cor. 8.7 is generally false: for example, if $ \forall n \in \mathbb{n}, u_n=(-1)^n$ , then $ (\vert u_n\vert)$ is constant, thus convergent, but $ (u_n)$ is periodic with period 2, thus divergent.

For $ l=0$ , the converse of Cor. 8.7 is true.

Proposition 3.8.8   Let $ (u_n)$ be a sequence of real numbers.
  1. If $ (u_n)$ is an increasing sequence and is bounded above, then it is convergent.
  2. If $ (u_n)$ is a dereasing sequence and is bounded below, then it is convergent.

Example 3.8.9   Let $ (u_n)$ be defined by \begin{displaymath}\begin{cases}u_0=1  u_{n+1}=\sqrt{2+u_n} \end{cases}\end{displaymath} . Thus, $ (u_n)$ is convergent.

Noah Dana-Picard 2007-12-28