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

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## Contents

## 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:

\(\displaystyle x_2\) | \(=\) | \(\displaystyle \dfrac k {1 + x_1}\) | |||||||||||

\(\displaystyle \) | \(<\) | \(\displaystyle \dfrac k 1\) | as $x_1 > 0$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle k\) |

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:

\(\displaystyle x_{r + 1}\) | \(=\) | \(\displaystyle \dfrac k {1 + x_r}\) | |||||||||||

\(\displaystyle \) | \(<\) | \(\displaystyle \dfrac k 1\) | Induction Hypothesis: $x_r > 0$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle k\) |

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

- 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 5$: Subsequences: Exercise $\S 5.7 \ (3)$