Convergent Real Sequence/Examples

From ProofWiki
Jump to navigation Jump to search

Examples of Convergent Real Sequences

Example: $1 + \dfrac 1 n$

The sequence $\sequence {a_n}_{n \mathop \ge 1}$ defined as:

$a_n := 1 + \dfrac 1 n$

is convergent to the limit $1$ as $n \to \infty$.


Example: $\dfrac {n^2 - 1} {n^2 + 1}$

$\ds \lim_{n \mathop \to \infty} \paren {\dfrac {n^2 - 1} {n^2 + 1} } = 1$


Example: $\dfrac {2 n^3 - 3 n} {5 n^3 + 4 n^2 - 2}$

$\ds \lim_{n \mathop \to \infty} \paren {\dfrac {2 n^3 - 3 n} {5 n^3 + 4 n^2 - 2} } = \dfrac 2 5$


Example: $\dfrac {n^3 + 5 n^2 + 2} {2 n^3 + 9}$

$\ds \lim_{n \mathop \to \infty} \paren {\dfrac {n^3 + 5 n^2 + 2} {2 n^3 + 9} } = \dfrac 1 2$


Example: $\dfrac {x + x^n} {1 + x^n}$

The sequence $\sequence {a_n}$ defined as:

$a_n = \dfrac {x + x^n} {1 + x^n}$

is convergent for $x \ne -1$.

Then:

$\ds \lim_{n \mathop \to \infty} \dfrac {x + x^n} {1 + x^n} = \begin {cases} 1 & : x \ge 1 \\ x & : -1 < x < 1 \\ 1 & : x < -1 \\ \text {undefined} & : x = -1

\end {cases}$


Example: $x_{n + 1} = {x_n}^2 + k$

Let $\sequence {x_n}$ be the real sequence defined as:

$x_n = \begin {cases} h & : n = 1 \\ {x_{n - 1} }^2 + k & : n > 1 \end {cases}$

where:

$0 < k < \dfrac 1 4$
$a < h < b$, where $a$ and $b$ are the roots of the quadratic equation: $x^2 - x + k = 0$.


Then $\sequence {x_n}$ is convergent such that:

$\ds \lim_{n \mathop \to \infty} x_n = a$


Example: $x_{n + 1} = \dfrac k {1 + x_n}$

Let $h, k \in \R_{>0}$.

Let $\sequence {x_n}$ be the real sequence defined as:

$x_n = \begin {cases} h & : n = 1 \\ \dfrac k {1 + x_{n - 1} } & : n > 1 \end {cases}$


Then $\sequence {x_n}$ is convergent to the positive root of the quadratic equation:

$x^2 + x = k$


Example: $x_n = \sqrt {x_{n - 1} y_{n - 1} }$, $\dfrac 1 {y_n} = \dfrac 1 2 \paren {\dfrac 1 {x_n} + \dfrac 1 {y_{n - 1} } }$

Let $\sequence {x_n}$ and $\sequence {y_n}$ be the real sequences defined as:

$x_n = \begin {cases} \dfrac 1 2 & : n = 1 \\ \sqrt {x_{n - 1} y_{n - 1} } & : n > 1 \end {cases}$
$\dfrac 1 {y_n} = \begin {cases} 1 & : n = 1 \\ \dfrac 1 2 \paren {\dfrac 1 {x_n} + \dfrac 1 {y_{n - 1} } } & : n > 1 \end {cases}$

Then both $\sequence {x_n}$ and $\sequence {y_n}$ converge to the limit $\dfrac \pi 4$.


Example: $x_n = \dfrac {x_{n - 1} + x_{n - 2} } 2$

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.


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

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.


Example: $x_n = \sqrt {x_{n - 1} x_{n - 2} }$

Let $a, b \in \R_{>0}$ be (strictly) positive real numbers such that $a \le b$.

Let $a, b \in \R_{>0}$ be such that $a \le p \le q \le b$.

Let $\sequence {x_n}_{n \mathop \in \N_{>0} }$ be the sequence in $\R$ defined as:

$x_n = \begin {cases} p & : n = 1 \\ q & : n = 2 \\ \sqrt{x_{n - 1} x_{n - 2} } & : n > 2 \end {cases}$

That is, beyond the first $2$ terms, each term is the geometric mean of the previous $2$ terms.

Then $\sequence {x_n}$ converges.