# Convergent Real Sequence/Examples/Difference between Adjacent Terms Bounded by alpha^n

## Example of Convergent Real Sequence

Let $\alpha \in \R$ be a real number such that $0 < \alpha < 1$.

Let $\sequence {x_n}_{n \mathop \in \N_{>0} }$ be a sequence in $\R$ with the property:

$\size {x_{n + 1} - x_n} \le \alpha^n$

Then $\sequence {x_n}$ is a Cauchy sequence and hence converges.

## Proof

Let $n > m$.

Then:

 $\displaystyle \size {x_n - x_m}$ $=$ $\displaystyle \size {\paren {x_n - x_{n - 1} } + \paren {x_{n - 1} - x_{n - 2} } + \dotsb + \paren {x_{m + 1} - x_m} }$ $\displaystyle$ $\le$ $\displaystyle \size {x_n - x_{n - 1} } + \size {x_{n - 1} - x_{n - 2} } + \dotsb + \size {x_{m + 1} - x_m}$ Triangle Inequality for Real Numbers $\displaystyle$ $\le$ $\displaystyle \alpha^{n - 1} + \alpha^{n - 2} + \dotsb + \alpha^m$ Definition of $\sequence {x_n}$ $\displaystyle$ $=$ $\displaystyle \alpha^m \paren {\dfrac {1 - \alpha^{n - m} } {1 - \alpha} }$ Sum of Geometric Progression $\displaystyle$ $<$ $\displaystyle \dfrac {\alpha^m} {1 - \alpha}$

Let $\epsilon \in \R_{>0}$ be given.

Let $N$ be sufficiently large that:

$\dfrac {\alpha^m} {1 - \alpha} < \epsilon$

This is always possible, as shown by Sequence of Powers of Number less than One.

Hence the result, by Cauchy Sequence Converges on Real Number Line.

$\blacksquare$