Convergent Real Sequence/Examples/n^2 - 1 over n^2 + 1

Example of Convergent Real Sequence

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

Let $\epsilon \in \R_{>0}$ be given.

The requirement is to find a value of $N \in \R$ such that:

$\forall n > N: \size {\dfrac {n^2 - 1} {n^2 + 1} - 1} < \epsilon$

But we have:

$\ds \size {\dfrac {n^2 - 1} {n^2 + 1} - 1}$

$=$

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

$\ds $

$=$

$\ds \dfrac 2 {n^2 + 1}$

Hence the requirement is now to find a value of $N \in \R$ such that:

$\forall n > N: \dfrac 2 {n^2 + 1} < \epsilon$

$\dfrac 2 {n^2 + 1} < \epsilon \implies n^2 + 1 > \dfrac 2 \epsilon$

So choosing $N = \paren {\dfrac 2 \epsilon}^{1/2}$ we have that:

$\forall n > N: n > \dfrac 1 \epsilon$

$\ds \forall n > N: \, $

$\ds n^2 + 1$

$>$

$\ds n^2$

$\ds $

$>$

$\ds N^2$

$\ds $

$=$

$\ds \dfrac 2 \epsilon$

We have demonstrated that for $\epsilon \in \R_{>0}$ there exists $N \in \R$, that is: $N = \paren {\dfrac 2 \epsilon}^{1/2}$, such that:

$\forall n > N: \size {\dfrac {n^2 - 1} {n^2 + 1} - 1} < \epsilon$

Hence the result:

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

$\blacksquare$