Definition and examples.

Definition 9.1.1   Let be given a sequence $ (u_n)_{n \geq 0}$ of real numbers. The expression

$\displaystyle \underset{k=0}{\overset{+\infty }{\sum}} u_k =u_0+u_1+u_2+ \dots + U_n + \dots$    

is called the infinite series with $ u_n$ as its general term.

Definition 9.1.2   The sequence $ (S_n)_{n \geq 0}$ whose general term is $ S_n=\underset{k=0}{\overset{n}{\sum}} u_k = u_0+u_1 + \dots + u_n$ is called the sequence of partial sums of the given series.

Definition 9.1.3   The series $ \underset{k=0}{\overset{+\infty }{\sum}} u_k$ is convergent if its sequence of partial sums is convergent. Otherwise it is divergent.

if the sequence of partial sums converges, its limit is called the sum of the series.

For the definition of the convergence of a sequence, v.s. Def. 6.1.

Proposition 9.1.4 (Geometric Series)   A geometric series is a series $ \underset{k=0}{\overset{+\infty }{\sum}} u_k$ , where $ (u_n)_{n \geq 0}$ is a geometric sequence. The ratio of the sequence $ (u_n)_{n \geq 0}$ is also called the ratio of the geometric series.

Denote $ u_0 \neq 0$ the first term of the geometric series and by $ q$ its ratio; according to Prop. 4.2, we have: $ S_n= \underset{k=0}{\overset{n}{\sum}} u_k= u_0 \frac {1-q^{n+1}}{1-q}$ . thus:

Example 9.1.5   Let $ x=0.999999...$ , i.e. $ x= 9 \cdot 10^{-1} + 9 \cdot 10^{-3}+ 9 \cdot 10^{-3}+ \dots$ . The number x is the sum of a geometric series, whose first term is equal to $ 9 \cdot 10^{-1}$ and whose ratio is equal to $ 10^{-1}$ . Thus:

$\displaystyle x= 9 \cdot 10^{-1} \frac {1}{1-10^{-1}} = 1.$    

Example 9.1.6 (A Telescoping Series)   Consider the series whose general term is $ u_n=\frac{1}{n(n+1}$ . we have:

$\displaystyle \forall k \in \mathbb{N}-{0}, \; u_k= \frac 1k - \frac {1}{k+1}.$    

Hence:

$\displaystyle \underset{k=0}{\overset{n}{\sum}} u_k$ $\displaystyle =\underset{k=0}{\overset{+\infty }{\sum}} \left(\frac 1k - \frac {1}{k+1} \right)$    
$\displaystyle \quad$ $\displaystyle =\frac 11 - \frac 12 + \frac 12 - \frac 13 + \frac 13 - \frac 14 + \dots + \frac 1n - \frac {1}{n+1}$    
$\displaystyle \quad$ $\displaystyle 1 - \frac {1}{n+1}$    

It follows that the series $ \underset{k=0}{\overset{+\infty }{\sum}} \frac {1}{k(k+1)}$ . is convergent and its sum is equal to 1.

Proposition 9.1.7   We consider the series $ \underset{n=0}{\overset{+\infty }{\sum}} u_n$ . If it is convergent, then $ \underset{n \rightarrow + \infty}{\text{lim}} u_n =0$ .

Example 9.1.8   Once again, consider the telescopic series from Example 1.6. We have: $ \underset{n \rightarrow + \infty}{\text{lim}} \frac {1}{n(n+1)}=0$ .

\begin{figure}\mbox{\subfigure{\epsfig{file=error.eps, height=1cm}} \qquad
The converse is not true!}
\end{figure}

Take the series:

$\displaystyle 1+\underbrace{\frac 12 + \frac 12 }_{=1} + \underbrace{\frac 13 +...
...rac 13 }_{=1} +\underbrace{\frac 14 +\frac 14 +\frac 14 +\frac 14 }_{=1}+ \dots$    

Despite the fact that the general term of the series has a limit equal to 0, this series is divergent (it has an infinite sum).

Example 9.1.9 (The harmonic series)   Take the series $ \underset{k=1}{\overset{+ \infty}{\sum}} \frac 1k = 1+ \frac 12 \frac 13 + \frac 14 + \dots$ .

We prove that this series is divergent by grouping the terms in the following way:

$\displaystyle 1+ \frac 12 \underbrace{ \frac 13 + \frac 14}_{> \frac 12} +\underbrace{ \frac 15 + \frac 16 + \frac 17 + \frac 18}_{> \frac 12} + \dots$    

Noah Dana-Picard 2007-12-28