Convergent Real Sequence/Examples/Arithmetic Mean of Previous 2 Terms
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Example of Convergent Real Sequence
Let $\sequence {x_n}_{n \mathop \in \N_{>0} }$ be the sequence in $\R$ defined as:
- $x_n = \begin {cases} a & : n = 1 \\ b & : n = 2 \\ \dfrac {x_{n - 1} + x_{n - 2} } 2 & : n > 2 \end {cases}$
That is, beyond the first $2$ terms, each term is the arithmetic mean of the previous $2$ terms.
Then $\sequence {x_n}$ converges.
Proof
\(\ds x_{n + 2} - x_{n + 1}\) | \(=\) | \(\ds \dfrac {x_{n + 1} + x_n} 2 - x_{n + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {x_n - x_{n + 1} } 2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \size {x_{n + 2} - x_{n + 1} }\) | \(=\) | \(\ds \dfrac {\size {x_n - x_{n + 1} } } 2\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\size {x_n - x_{n - 1} } } {2^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \ldots\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\size {x_2 - x_1} } {2^n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\size {b - a} } {2^n}\) |
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 | |||||||||||
\(\ds \) | \(\le\) | \(\ds \paren {\dfrac 1 {2^{n - 2} } + \dfrac 1 {2^{n - 3} } + \dotsb + \dfrac 1 {2^{m - 1} } } \size {b - a}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {2^{m - 1} } \paren {1 + \dfrac 1 2 + \dfrac 1 {2^2} + \dotsb + \dfrac 1 {2^{n - m - 1} } } \size {b - a}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {2^{m - 1} } \paren {\dfrac {1 - \paren {\frac 1 2}^{n - m} } {1 - \frac 1 2} } \size {b - a}\) | Sum of Geometric Sequence | |||||||||||
\(\ds \) | \(\le\) | \(\ds \dfrac 1 {2^{m - 2} } \size {b - a}\) |
Let $\epsilon \in \R_{>0}$ be given.
Let $N$ be sufficiently large that:
- $\dfrac 1 {2^{N - 2} } \size {b - a} < \epsilon$
Then:
- $\forall n > N, m > N: \size {x_n - x_m} \le \dfrac 1 {2^{N - 2} } \size {b - a} < \epsilon$
Hence it is seen that $\sequence {x_n}$ is a Cauchy sequence.
Hence the result, by Cauchy's Convergence Criterion.
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 5$: Subsequences: Cauchy sequences: Example $\S 5.20$