Convergent Real Sequence/Examples/x (n+1) = k over 1 + x n/Lemma 1

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Example of Convergent Real Sequence

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:

$\forall n \in \N_{>1}: k > x_n > 0$


Proof

The proof proceeds by induction.

For all $n \in \Z_{>1}$, let $\map P n$ be the proposition:

$k > x_n > 0$


Basis for the Induction

$\map P 2$ is the case:

$k > x_2 > 0$

We have:

\(\ds x_2\) \(=\) \(\ds \dfrac k {1 + x_1}\)
\(\ds \) \(<\) \(\ds \dfrac k 1\) as $x_1 > 0$
\(\ds \) \(=\) \(\ds k\)

Also, as $k > 0$ and $x_1 > 0$ we have that:

$\dfrac k {1 + x_1} > 0$


Thus $\map P 2$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that if $\map P r$ is true, where $r \ge 2$, then it logically follows that $\map P {r + 1}$ is true.


So this is the induction hypothesis:

$k > x_r > 0$


from which it is to be shown that:

$k > x_{r + 1} > 0$


Induction Step

This is the induction step:

\(\ds x_{r + 1}\) \(=\) \(\ds \dfrac k {1 + x_r}\)
\(\ds \) \(<\) \(\ds \dfrac k 1\) Induction Hypothesis: $x_r > 0$
\(\ds \) \(=\) \(\ds k\)

Also, as $k > 0$ and $x_r > 0$ we have that:

$\dfrac k {1 + x_r} > 0$

So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n \in \N_{>1}: k > x_n > 0$

$\blacksquare$


Sources

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