# Convergent Real Sequence/Examples/x (n+1) = k over 1 + x n/Lemma 2

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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:

- $x_n = \begin {cases} h & : n = 1 \\ \dfrac k {1 + x_{n - 1} } & : n > 1 \end {cases}$

Consider the subsequences $\sequence {x_{2 n} }$ and $\sequence {x_{2 n - 1} }$.

One of them is strictly increasing and the other is strictly decreasing.

## Proof

We have that:

\(\ds x_{n + 1} - x_{n - 1}\) | \(=\) | \(\ds \dfrac k {1 + x_n} - \dfrac k {1 + x_{n - 2} }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \dfrac {k \paren {x_{n - 2} - x_n} } {\paren {1 + x_n} \paren {1 + x_{n - 2} } }\) |

and so $x_{n + 1} - x_{n - 1}$ has the opposite sign to $x_{n - 2} - x_n$.

It can be proved by induction that one of the sequences $\sequence {x_{2 n} }$ and $\sequence {x_{2 n - 1} }$ increases and one decreases.

This needs considerable tedious hard slog to complete it.Provide the workings for the aboveTo discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Finish}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

In fact:

- $\sequence {x_{2 n - 1} }$ is strictly increasing if and only if $x_3 > x_1$ and is strictly decreasing if and only if $x_3 < x_1$.

This needs considerable tedious hard slog to complete it.Prove the above as well.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Finish}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

## Sources

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