Real Sequence/Examples/x(n+1) = 2 over x(n)+1

Example of Real Sequence
Let $\sequence {x_n}$ denote the real sequence defined as:
 * $x_n = \begin {cases} a : 0 < a < 1 & : n = 1 \\ \dfrac 2 {x_n + 1} & : n > 1 \end {cases}$

Then the subsequences $\sequence {x_{2 n} }$ and $\sequence {x_{2 n + 1} }$ are both monotone:


 * $\sequence {2 n}$ is strictly decreasing
 * $\sequence {2 n + 1}$ is strictly increasing

Hence $\sequence {x_n} \to 1$ as $n \to \infty$.

Proof
We are given that $0 < x_1 < 1$.

Then we have that:
 * $\dfrac 2 {0 + 1} > x_2 > \dfrac 2 {1 + 1}$

That is:
 * $1 < x_2 < 2$

Then we have that:
 * $\dfrac 2 {1 + 1} > x_3 > \dfrac 2 {2 + 1}$

That is:
 * $\dfrac 2 3 < x_3 < 1$

and so certainly:
 * $0 < x_3 < 1$

Hence in general:
 * $0 < x_n < 1 \implies 1 < x_{n + 1} < 2$

and:
 * $1 < x_n < 2 \implies 0 < x_{n + 1} < 1$

So:
 * for $n$ odd, $0 < x_n < 1$
 * for $n$ even, $1 < x_n < 2$

Now we show:

We calculate the difference between $x_n$ and $x_{n + 2}$:

and we see that:
 * if $0 < x_n < 1$ then $x_n - x_{n + 2} < 0$
 * if $1 < x_n < 2$ then $x_n - x_{n + 2} > 0$

Hence it follows that:
 * for $n$ odd, $\sequence {2 n + 1}$ is strictly increasing
 * for $n$ even, $\sequence {2 n}$ is strictly decreasing

and the result follows from Convergence of Odd and Even Subsequences to Same Limit.