Convergence and Divergence.

Definition 3.6.1   The sequence $ (u_n)$ is convergent if there exists a real number $ l$ verifying the following property (called Cauchy's inequality):

$\displaystyle \forall \varepsilon >0, \exists N_0 \in \mathbb{N} \; \vert \; n \geq N_0 \Longrightarrow \begin{vmatrix}u_n -l \end{vmatrix} < \varepsilon$    

Otherwise, the sequence $ (u_n)$ is divergent.

Notation: $ \underset{n \rightarrow \infty}{\text{lim}} u_n = l$ .

Example 3.6.2   Let $ u_n=\frac {n-1}{2n+1}$ ; we prove that $ (u_n)$ is convergent and that $ \underset{n \rightarrow \infty}{\text{lim}} u_n = \frac 12$

Take any $ \varepsilon >0$ ; we look for natural numbers such that $ \begin{vmatrix}\frac {n-1}{2n+1} -\frac 12 \end{vmatrix} < \varepsilon$ . We have:

$\displaystyle \begin{vmatrix}\frac {n-1}{2n+1} -\frac 12 \end{vmatrix}$ $\displaystyle < \varepsilon$    
$\displaystyle \begin{vmatrix}\frac {2(n-1)}{2(2n+1)} -\frac {2n+1}{2(2n+1)} \end{vmatrix}$ $\displaystyle < \varepsilon$    
$\displaystyle \begin{vmatrix}\frac {-3}{2(2n+1)} \end{vmatrix}$ $\displaystyle < \varepsilon$    
$\displaystyle \frac {3}{4n+2}$ $\displaystyle < \varepsilon$    
$\displaystyle 4n+2$ $\displaystyle > \frac {3}{\varepsilon}$    
$\displaystyle n$ $\displaystyle > \frac 14 \left( \frac {3}{\varepsilon} -2 \right).$    

Let $ N_0$ be the integer part of $ \frac 14 \left( \frac {3}{\varepsilon} -2 \right)$ ; then for any natural $ n$ such that $ n>N_0$ , we have $ \vert u_n -1/2 \vert< \varepsilon$ .

Proposition 3.6.3   If a sequence $ (a_n)$ is convergent, then it is bounded.

Proof. Let $ a$ be the limit of the sequence $ (a_n)$ ; by definition 6.1 we have:

$\displaystyle \forall \varepsilon >0, \exists N_0 \in \mathbb{N} \; \vert \; n \geq N_0 \Longrightarrow \begin{vmatrix}a_n -a \end{vmatrix} < \varepsilon .$    

Thus the set $ \{ a_n \; \vert \; n >N_0 \}$ is bounded (below by $ a -\varepsilon$ and above by $ a+ \varepsilon$ ). As the set $ \{ a_n \; \vert \; n \leq N_0 \}$ is finite, it is bounded. The set of all the terms of the sequence $ (a_n)$ is the union of two bounded sets, hence it is bounded. $ \qedsymbol$

\begin{figure}\mbox{\subfigure{\epsfig{file=error.eps, height=1cm}} \qquad
The converse is not true!}
\end{figure}
For example, any periodic sequence is bounded, but is divergent.

Theorem 3.6.4   Let $ (u_n)$ be an arithmetic sequence whose difference is equal to $ d$ .
  1. If $ d=0$ , the sequence is constant.
  2. If $ d>0$ , then the sequence increases and has $ +\infty$ as its limit.
  3. If $ d<0$ , then the sequence increases and has $ -\infty $ as its limit.

Theorem 3.6.5   Let $ (v_n)$ be an geometric sequence whose quotient is equal to $ q$ .
  1. If $ q=1$ , the sequence is constant.
  2. If $ q=0$ , the sequence is constant from its second term.
  3. If $ \vert q\vert<1$ , the sequence is convergent and its limit is 0.
  4. If $ q>1$ , the sequence is divergent; its limit is infinite ($ +\infty$ or $ -\infty $ according to the sign of the first term).
  5. If $ q<-1$ , the sequence is divergent and has no limit.
  6. If $ q=-1$ , the sequence is periodic, with period equal to 2.

Definition 3.6.6   The sequence $ (u_n)$ has $ +\infty$ as its limit if:

$\displaystyle \forall A>0, \exists N_0 \in \mathbb{N} \; \vert \; n \geq N_0 \Longrightarrow u_n >A$    

We denote: $ \underset{n \rightarrow \infty}{\text{lim}}u_n = +\infty$ .

This formula is called Cauchy's inequality too. Write the corresponding definition for $ \underset{n \rightarrow \infty}{\text{lim}}u_n=-\infty$ .

Noah Dana-Picard 2007-12-28