Definition:Convergent Sequence/Real Numbers

From ProofWiki
Jump to navigation Jump to search

Definition

Let $\sequence {x_k}$ be a sequence in $\R$.

The sequence $\sequence {x_k}$ converges to the limit $l \in \R$ if and only if:

$\forall \epsilon \in \R_{>0}: \exists N \in \R_{>0}: n > N \implies \size {x_n - l} < \epsilon$

where $\size x$ denotes the absolute value of $x$.


Graphical Illustration

The following diagram illustrates the first few terms of a convergent real sequence.


Convergent-real-sequence.png


For the value of $\epsilon$ given, a suitable value of $N$ such that:

$n > N \implies \size {x_n - l} < \epsilon$

is $6$.


Note on Domain of $N$

Some sources insist that $N \in \N$ but this is not strictly necessary and can make proofs more cumbersome.


Examples

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

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


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

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


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

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


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:

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


Also see


Generalizations


Sources