# Geometric series/Ratio test/R/Section

The series ${\displaystyle {}\sum _{k=0}^{\infty }x^{k}}$ is called geometric series for ${\displaystyle {}x\in \mathbb {R} }$, so this is the sum

${\displaystyle 1+x+x^{2}+x^{3}+\ldots .}$

The convergence depends heavily on the modulus of ${\displaystyle {}x}$.

## Theorem

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

${\displaystyle {}\sum _{k=0}^{\infty }x^{k}={\frac {1}{1-x}}\,.}$

### Proof

For every ${\displaystyle {}x}$ and every ${\displaystyle {}n\in \mathbb {N} }$ we have the relation

${\displaystyle {}(x-1){\left(\sum _{k=0}^{n}x^{k}\right)}=x^{n+1}-1\,}$

and hence for the partial sums the relation (for ${\displaystyle {}x\neq 1}$ )

${\displaystyle {}s_{n}=\sum _{k=0}^{n}x^{k}={\frac {x^{n+1}-1}{x-1}}\,}$

holds. For ${\displaystyle {}n\rightarrow \infty }$ and ${\displaystyle {}\vert {x}\vert <1}$ this converges to ${\displaystyle {}{\frac {-1}{x-1}}={\frac {1}{1-x}}}$ because of fact and exercise.

${\displaystyle \Box }$

The following statement is called ratio test.

## Theorem

Let

${\displaystyle \sum _{k=0}^{\infty }a_{k}}$

be a series of real numbers. Suppose there exists a real number ${\displaystyle {}q}$ with ${\displaystyle {}0\leq q<1}$, and a ${\displaystyle {}k_{0}}$ with

${\displaystyle {}\vert {\frac {a_{k+1}}{a_{k}}}\vert \leq q\,}$

for all ${\displaystyle {}k\geq k_{0}}$ (in particular ${\displaystyle {}a_{k}\neq 0}$ for ${\displaystyle {}k\geq k_{0}}$). Then the series ${\displaystyle {}\sum _{k=0}^{\infty }a_{k}}$ converges absolutely.

### Proof

The convergence does not change (though the sum) when we change finitely many members of the series. Therefore, we can assume ${\displaystyle {}k_{0}=0}$. Moreover, we can assume that all ${\displaystyle {}a_{k}}$ are positive real numbers. Then

${\displaystyle {}a_{k}={\frac {a_{k}}{a_{k-1}}}\cdot {\frac {a_{k-1}}{a_{k-2}}}\cdots {\frac {a_{1}}{a_{0}}}\cdot a_{0}\leq a_{0}\cdot q^{k}\,.}$

Hence, the convergence follows from the comparison test and the convergence of the geometric series.

${\displaystyle \Box }$

## Example

The Koch snowflakes are given by the sequence of plane geometric shapes ${\displaystyle {}K_{n}}$, which are defined recursively in the following way: The starting object ${\displaystyle {}K_{0}}$ is an equilateral triangle. The object ${\displaystyle {}K_{n+1}}$ is obtained from ${\displaystyle {}K_{n}}$ by replacing in each edge of ${\displaystyle {}K_{n}}$ the third in the middle by the corresponding equilateral triangle showing outside.

Let ${\displaystyle {}A_{n}}$ denote the area and ${\displaystyle {}L_{n}}$ the length of the boundary of the ${\displaystyle {}n}$-th Koch snowflake. We want to show that the sequence ${\displaystyle {}A_{n}}$ converges and that the sequence ${\displaystyle {}L_{n}}$ diverges to ${\displaystyle {}\infty }$.

The number of edges of ${\displaystyle {}K_{n}}$ is ${\displaystyle {}3\cdot 4^{n}}$, since in each division step, one edge is replaced by four edges. Their length is ${\displaystyle {}1/3}$ of the length of a previous edge. Let ${\displaystyle {}r}$ denote the base length of the starting equilateral triangle. Then ${\displaystyle {}K_{n}}$ consists of ${\displaystyle {}3\cdot 4^{n}}$ edges of length ${\displaystyle {}r{\left({\frac {1}{3}}\right)}^{n}}$ and the length of all edges of ${\displaystyle {}K_{n}}$ together is

${\displaystyle {}L_{n}=3\cdot 4^{n}r{\left({\frac {1}{3}}\right)}^{n}=3r{\left({\frac {4}{3}}\right)}^{n}\,.}$

Because of ${\displaystyle {}{\frac {4}{3}}>1}$, this diverges to ${\displaystyle {}\infty }$.

When we turn from ${\displaystyle {}K_{n}}$ to ${\displaystyle {}K_{n+1}}$, 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 ${\displaystyle {}s}$ is ${\displaystyle {}{\frac {\sqrt {3}}{4}}s^{2}}$. So in the step from ${\displaystyle {}K_{n}}$ to ${\displaystyle {}K_{n+1}}$ there are ${\displaystyle {}3\cdot 4^{n}}$ triangles added with area ${\displaystyle {}{\frac {\sqrt {3}}{4}}{\left({\frac {1}{3}}\right)}^{2(n+1)}r^{2}={\frac {\sqrt {3}}{4}}r^{2}{\left({\frac {1}{9}}\right)}^{n+1}}$. The total area of ${\displaystyle {}K_{n}}$ is therefore

{\displaystyle {}{\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 ${\displaystyle {}1}$ and the factor ${\displaystyle {}{\frac {3}{4}}}$, which does not change the convergence property, we get in the bracket a partial sum of the geometric series for ${\displaystyle {}{\frac {4}{9}}}$, and this converges.