I have the sequence
$$
x_{n+1} = 2^{n+1} \sqrt{1 + 2^{-n} x_n} – 1,
\quad \mbox{for}~n\geq 1.
$$
We assume $x_0 > – 1$.
I need to prove that the sequence $\{ x_n \}_{n \geq 0}$ is bounded, monotone, and convergent, with limit
$$
\ln (x_0 + 1)
$$
The text gave the hint to rationalize, so I got to $x_{n+1} = \frac{2* x_n}{\sqrt{1+2^{-n}x_n}+1}$. I ran some calculations and it is monotone decreacing, but I can't seem to get how to prove it.
Any help would be appreciated.
Best Answer
We can write $$ x_{n+1} = 2^{n+1} \Big(\sqrt{1 + 2^{-n} x_n} - 1\Big) $$ Let us then define the auxiliary sequence $y_n = 2^{-n} x_n + 1$. We have $$ y_{n+1} = \sqrt{y_n}. $$ Therefore, $y_n = y_0^{1/(2^n)}$. Note $y_0 = x_0 + 1$. Consequently, $$ \lim x_n = \lim 2^n(y_n - 1)= \lim 2^n\Big[\exp\Big(\frac{1}{2^n} \log(1 + x_0)\Big) - 1\Big] = \log( 1 + x_0), $$ by L'Hopital's rule.
That $x_n$ is bounded now follows trivially from the limit relation. As for the monotonicity, we start by defining the function $$ \psi(t) = t \Big[\exp\Big(\frac{c}{t}\Big) - 1\Big], \quad \mbox{where}~c = \log(1+x_0). $$ Then $$ \psi'(t) = e^{c/t} \Big(1 - \frac{c}{t}\Big) - 1 \leq 0. $$ where above we use the inequality $e^x \geq 1 + x$ for all real $x$. This establishes that $\psi$ is a non increasing function. Now we note that $$ x_n = \psi(2^n). $$ Then $x_{n+1} = \psi(2^{n+1}) \leq \psi(2^n) = x_n$. Note that this provides yet another proof of the boundedness of $x_n$, since $$ \sup_n x_n = x_0, \quad \mbox{and} \quad \inf_n x_n = \lim_n x_n = \log(1 + x_0). $$