# Geometric series/Ratio test/R/Section

The series $\sum _{k=0}^{\infty }x^{k}$ is called geometric series for $x\in \mathbb {R}$, so this is the sum
    $1+x+x^{2}+x^{3}+\ldots .$
The convergence depends heavily on the modulus of $x$.

## Theorem

For all real numbers $x$ with $\vert {x}\vert <1$, the geometric series $\sum _{k=0}^{\infty }x^{k}$ converges absolutely, and the sum equals
    $\sum _{k=0}^{\infty }x^{k}={\frac {1}{1-x}}\,.$

### Proof

For every $x$ and every $n\in \mathbb {N}$ we have the relation
    $(x-1){\left(\sum _{k=0}^{n}x^{k}\right)}=x^{n+1}-1\,$
and hence for the partial sums the relation (for $x\neq 1$ )
    $s_{n}=\sum _{k=0}^{n}x^{k}={\frac {x^{n+1}-1}{x-1}}\,$
holds. For $n\rightarrow \infty$ and $\vert {x}\vert <1$ this converges to ${\frac {-1}{x-1}}={\frac {1}{1-x}}$ because of fact and exercise.
$\Box$
The following statement is called ratio test.

## Theorem

Let
    $\sum _{k=0}^{\infty }a_{k}$
be a series of real numbers. Suppose there exists a real number $q$ with $0\leq q<1$, and a $k_{0}$ with
    $\vert {\frac {a_{k+1}}{a_{k}}}\vert \leq q\,$
for all $k\geq k_{0}$ (in particular $a_{k}\neq 0$ for $k\geq k_{0}$). Then the series $\sum _{k=0}^{\infty }a_{k}$

### Proof

The convergence does not change (though the sum) when we change finitely many members of the series. Therefore, we can assume $k_{0}=0$. Moreover, we can assume that all $a_{k}$ are positive real numbers. Then
    $a_{k}={\frac {a_{k}}{a_{k-1}}}\cdot {\frac {a_{k-1}}{a_{k-2}}}\cdots {\frac {a_{1}}{a_{0}}}\cdot a_{0}\leq a_{0}\cdot q^{k}\,.$
Hence, the convergence follows from the comparison test and the convergence of the geometric series.
$\Box$

## Example

The Koch snowflakes are given by the sequence of plane geometric shapes $K_{n}$, which are defined recursively in the following way: The starting object $K_{0}$ is an equilateral triangle. The object $K_{n+1}$ is obtained from $K_{n}$ by replacing in each edge of $K_{n}$ the third in the middle by the corresponding equilateral triangle showing outside.
Let $A_{n}$ denote the area and $L_{n}$ the length of the boundary of the $n$-th Koch snowflake. We want to show that the sequence $A_{n}$ converges and that the sequence $L_{n}$ diverges to $\infty$.
The number of edges of $K_{n}$ is $3\cdot 4^{n}$, since in each division step, one edge is replaced by four edges. Their length is $1/3$ of the length of a previous edge. Let $r$ denote the base length of the starting equilateral triangle. Then $K_{n}$ consists of $3\cdot 4^{n}$ edges of length $r{\left({\frac {1}{3}}\right)}^{n}$ and the length of all edges of $K_{n}$ together is
    $L_{n}=3\cdot 4^{n}r{\left({\frac {1}{3}}\right)}^{n}=3r{\left({\frac {4}{3}}\right)}^{n}\,.$
Because of ${\frac {4}{3}}>1$, this diverges to $\infty$.
When we turn from $K_{n}$ to $K_{n+1}$, there will be for every edge a new triangle whose side length is a third of the edge length. The area of an equilateral triangle with side length $s$ is ${\frac {\sqrt {3}}{4}}s^{2}$. So in the step from $K_{n}$ to $K_{n+1}$ there are $3\cdot 4^{n}$ triangles added with area ${\frac {\sqrt {3}}{4}}{\left({\frac {1}{3}}\right)}^{2(n+1)}r^{2}={\frac {\sqrt {3}}{4}}r^{2}{\left({\frac {1}{9}}\right)}^{n+1}$. The total area of $K_{n}$ is therefore
    ${\begin{aligned}&\,{\frac {\sqrt {3}}{4}}r^{2}{\left(1+3{\frac {1}{9}}+12{\left({\frac {1}{9}}\right)}^{2}+48{\left({\frac {1}{9}}\right)}^{3}+\cdots +3\cdot 4^{n-1}{\left({\frac {1}{9}}\right)}^{n}\right)}\\&={\frac {\sqrt {3}}{4}}r^{2}{\left(1+{\frac {3}{4}}{\left({\frac {4}{9}}\right)}^{1}+{\frac {3}{4}}{\left({\frac {4}{9}}\right)}^{2}+{\frac {3}{4}}{\left({\frac {4}{9}}\right)}^{3}+\cdots +{\frac {3}{4}}{\left({\frac {4}{9}}\right)}^{n}\right)}.\end{aligned}}$
If we forget the $1$ and the factor ${\frac {3}{4}}$, which does not change the convergence property, we get in the bracket a partial sum of the geometric series for ${\frac {4}{9}}$, and this converges.