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\title{Session 09 -- Geometric sequences}
\subtitle{European section -- Season 2}
\date{}

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  \titlepage
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\section{Definition and criterion}

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\txt{A sequence of numbers $(b_n)$ is geometric if the quotient
between two consecutive terms is a constant number. Intuitively, to
go from one term to the next one, we always multiply by the same
number.}
\pause

\defi{Geometric sequence}{A sequence of numbers $(b_n)$ is geometric if, for any
positive integer $n$, $\frac{b_{n+1}}{b_n}=q$ where $q$ is a fixed
real number, called the \emph{common ratio} of the
sequence. We can also write that $b_{n+1}=b_n\times q$. This
equality is called the \emph{recurrence relation} of the sequence.}

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\section{Relations between terms}

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\prop{Explicit definition}{For any positive integer $n$, $b_n=b_1\times q^{n-1}$. This
equality is called the \emph{explicit definition} of the
sequence.}\pause

\dem{First, this equality is true when $n=0$, as $$b_1=b_1\times
q^0=b_1\times q^{1-1}.$$\pause Now, suppose that it is true for a
value $n=k$, meaning that $b_k=b_1\times q^{k-1}$.\pause\ Then, from
the definition of the sequence, $$b_{k+1}=b_k\times q=b_1\times
q^{k-1}\times q=b_1+\times q^k.$$\pause\ So the formula is true for
$n=k+1$ too.\pause\ So it's true for $n=0$, $n=1$, $n=2$, $n=3$,
etc, for all values of $n$.}

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\prop{Relation between two terms}{For any two positive integers $n$ and $m$, $b_n=b_m\times
q^{n-m}$.}\pause

\dem{From the explicit definition of the sequence $(b_n)$,
$b_n=b_1\times q^{n-1}$ and $b_m=b_1\times q^{m-1}$, so\pause
$$\frac{b_n}{b_m}=\frac{b_1\times q^{n-1}}{b_m=b_1\times
q^{m-1}}=\frac{q^{n-1}}{q^{m-1}}=q^{n-m}.$$\pause Therefore,
$b_n=b_m\times q^{n-m}$.}

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\section{Limit when $n$ approaches $+\infty$}

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\theo{Limit of a geometric sequence}{The limit of a geometric sequence $(b_n)$ of common ratio
$q$ and first term $b_1$\pause
\begin{itemize}
\item is equal to $0$, trivially, if $b_1=0$ ;\pause
\item is equal to $b_1$, trivially, if $r=1$ ;\pause
\item is equal to $+\infty$ when $b_1>0$ and $q>1$ ;\pause
\item is equal to $-\infty$ when $b_1<0$ and $q>1$ ;\pause
\item is equal to $0$ if $q\in]-1;1[$ ;\pause
\item doesn't exist when $r\leq -1$.
\end{itemize}\pause
In the last situation, the sequence is said to be \emph{divergent}.}

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\dem{\begin{itemize}
\item If $b_1=0$ or if $r=1$, the result is obvious : in both
cases, the sequence is constant !
\end{itemize}}
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\dem{\begin{itemize}
\item If $b_1>0$ and $q>1$, consider any real number $K$. The
inequation $b_n>K$, or $b_1\times q^{n-1}>K$ is equivalent to
\begin{eqnarray*}
q^{n-1} &>& \frac{K}{b_1}\\
\ln(q^{n-1}) &>& \ln\left(\frac{K}{b_1}\right)\\
(n-1)\ln q &>& \ln K-\ln b_1\\
(n-1) &>& \frac{\ln K-\ln b_1}{\ln q}\\
n &>& \frac{\ln K-\ln b_1}{\ln q}+1
\end{eqnarray*}
This means that for any real number $K$, there exist some integer
$N$ such that for any $n\geq N$, $b_N>K$. This is exactly the
definition of the fact that $\lim b_n=+\infty$.
\end{itemize}}
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\begin{frame}
\dem{\begin{itemize}
\item If $b_1<0$ and $q>1$, consider any real number $K$. The
inequation $b_n<K$, or $b_1\times q^{n-1}<K$ is equivalent to
\begin{eqnarray*}
q^{n-1} &>& \frac{K}{b_1}\\
\ln(q^{n-1}) &>& \ln\left(\frac{K}{b_1}\right)\\
(n-1)\ln q &>& \ln K-\ln b_1\\
(n-1) &>& \frac{\ln K-\ln b_1}{\ln q}\\
n &>& \frac{\ln K-\ln b_1}{\ln q}+1
\end{eqnarray*}
This means that for any real number $K$, there exist some integer
$N$ such that for any $n\geq N$, $b_N<K$. This is exactly the
definition of the fact that $\lim b_n=-\infty$.
\end{itemize}}
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\begin{frame}
\dem{\begin{itemize}
\item If $q\in]-1;1[$, consider any positive number $\varepsilon$. The
inequation $|b_n|<\varepsilon$, or $|b_1|\times
|q|^{n-1}<\varepsilon$ is equivalent to
\begin{eqnarray*}
|q|^{n-1} &<& \frac{\varepsilon}{|b_1|}\\
\ln(|q|^{n-1}) &<& \ln\left(\frac{\varepsilon}{|b_1|}\right)\\
(n-1)\ln |q| &<& \ln \varepsilon-\ln |b_1|\\
(n-1) &>& \frac{\ln \varepsilon-\ln |b_1|}{\ln |q|}\\
n &>& \frac{\ln \varepsilon-\ln |b_1|}{\ln |q|}+1
\end{eqnarray*}
This means that for any positive real number $\varepsilon$, there
exists some integer $N$ such that for any $n\geq N$,
$|b_N|<\varepsilon$. This is exactly the definition of the fact that
$\lim b_n=0$.
\end{itemize}}
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\dem{\begin{itemize}
\item Finally when $r\leq -1$, the sequence is alternating between
positive and negative terms, whose absolute values approach
$+\infty$. So the sequence has no limit.
\end{itemize}}

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\section{Sums of consecutive terms}

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\theo{Sum of consecutive terms}{Let $(b_n)$ be an geometric sequence, The sum $S$ of the $n$ first consecutive terms,
defined as
$S=b_1+b_2+\ldots+b_{n-1}+b_n$, or more precisely
$S=\sum_{i=1}^{n} b_i$, is given by the formula \pause
$$S=b_1\frac{1-q^{n}}{1-q}.$$}

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\dem{Using the formula of proposition 2.2, we can write $S$ as 
\begin{eqnarray*}
S &=&\pause b_1+b_1\times q+\ldots+b_1\times q^{n-2}+b_1\times q^{n-1}\\ \pause
  &=& b_1\times(1+q+\ldots+q^{n-2}+q^{n-1}). \pause
\end{eqnarray*}
But, by a simple expansion, we see that
$(1+q+\ldots+q^{n-2}+q^{n-1})(1-q)=\pause1-q^{n}$ and so\pause
$$S=b_1\frac{1-q^{n}}{1-q}.$$}

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