Geometric series/Ratio test/R/Section

This image illustrates the behavior of the geometric series for ${}x={\frac {1}{4}}$. Suppose that the length of the square is ${}2$. Then we can put the geometric series threetimes inside this square. The area of the three series is ${}{\frac {4}{3}}$.

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

converges absolutely.

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.