# 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.In particular: 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.In particular: 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 $\S 5.7 \ (3)$