Convergent Real Sequence/Examples/Difference between Adjacent Terms Bounded by alpha^n
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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:
\(\ds \size {x_n - x_m}\) | \(=\) | \(\ds \size {\paren {x_n - x_{n - 1} } + \paren {x_{n - 1} - x_{n - 2} } + \dotsb + \paren {x_{m + 1} - x_m} }\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \size {x_n - x_{n - 1} } + \size {x_{n - 1} - x_{n - 2} } + \dotsb + \size {x_{m + 1} - x_m}\) | Triangle Inequality for Real Numbers | |||||||||||
\(\ds \) | \(\le\) | \(\ds \alpha^{n - 1} + \alpha^{n - 2} + \dotsb + \alpha^m\) | Definition of $\sequence {x_n}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \alpha^m \paren {\dfrac {1 - \alpha^{n - m} } {1 - \alpha} }\) | Sum of Geometric Sequence | |||||||||||
\(\ds \) | \(<\) | \(\ds \dfrac {\alpha^m} {1 - \alpha}\) |
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's Convergence Criterion.
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 5$: Subsequences: Exercise $\S 5.21 \ (1)$