# Show that $x_{n+1} = \frac{2x_n}{\sqrt{1+2^{-n}x_n}+1}$ is monotone and bounded

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.

• @dezdichado I thought so, but if $x_0 = 0.5$, $\sqrt{1-0.5} + 1 \approx 0.707 + 1 \approx 1.707$. Commented Oct 9, 2023 at 3:26

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 $$\sup_n x_n = x_0, \quad \mbox{and} \quad \inf_n x_n = \lim_n x_n = \log(1 + x_0).$$

• Excuse me, but I don't understand why can we assure that $y_{n+1} = \sqrt{y_n}$ ? Do the $-1$ and the $2^{-n}$ not change the recurrence relation? Commented Oct 9, 2023 at 4:18
• @Rararat: Rewrite the recurrence relation as $1 + 2^{-(n+1)}x_{n+1} = \sqrt{1 + 2^{-n} x_n}$ ... Commented Oct 9, 2023 at 4:40
• Thanks! And if I may ask, how do we get from $e^x \geq 1+x$ to $e^x * (1-x) -1 \leq 0$? I tried multiplying by $1-x$ both sides, using the fact that $x>0$ in this case, but I can't seem to get rid of the $x^2$ Commented Oct 9, 2023 at 20:11
• Apply the first inequality to $-x$. Does that help? Commented Oct 10, 2023 at 15:48