Showing that a sequence (defined in terms of the previous sequence term) is increasing and bounded above

Question

I'm stuck on this problem and I was wondering if you would be kind enough to help. The question follows:

Let $x_{1} = 1$ and $x_{n}$ = $\sqrt{ 1 + 2x_{n-1}}$ for n $\geq$ 2. Show that the sequence $\left ( x_{n} \right )_{n \geq 1}$ is increasing and bounded above.

We were told to use induction for the bounded above proof, which I have completed already by assuming that 3 is an upper bound of a certain value and then changing around the equations to show that the following sequence term is smaller than 3 also, so, I think I am done with that.

However, I'm stuck on proving that it is increasing for all values of n. Would you be able to help?

Thanks! Lauren

Answer 1

Let $f(x)=\sqrt{1+2x}$ then $x_n=f(x_{n-1})$ and since $f$ is an increasing function and $x_2>x_1$ then the sequence $(x_n)$ is increasing. Let $\ell=1+\sqrt 2$ the root greater than $1$ of the equation $f(x)=x$. Prove by induction that $x_n<\ell$.