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

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Example of Convergent Real Sequence

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


Proof

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:

\(\displaystyle \size {\dfrac {n^2 - 1} {n^2 + 1} - 1}\) \(=\) \(\displaystyle 1 - \dfrac {n^2 - 1} {n^2 + 1}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 2 {n^2 + 1}\)

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

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

From Reciprocal Function is Strictly Decreasing:

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

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

$\forall n > N: n > \dfrac 1 \epsilon$
\(\, \displaystyle \forall n > N: \, \) \(\displaystyle n^2 + 1\) \(>\) \(\displaystyle n^2\)
\(\displaystyle \) \(>\) \(\displaystyle N^2\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 2 \epsilon\)

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:

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

$\blacksquare$


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