$u_{n+2}=\frac{1}{2^{u_n}}+\frac{1}{2}$. find $\lim u_n$

Let ${u_n}$ be a real sequence defined by: $u_1=2$, $u_2=0$ $u_{n+2}=\frac{1}{2^{u_n}}+\frac{1}{2}$ Prove that ${u_n}$ has finite limit and find $\lim u_n$

Thanks

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This series is composed of 4 sub series, all of which are monotonic and bounded, try to show that, and show that all converge to the same limit. – yoki Nov 26 '12 at 20:24

HINT: Note first that the subsequences $\langle x_{2k}:k\in\Bbb Z^+\rangle$ and $\langle x_{2k-1}:k\in\Bbb Z^+\rangle$ are completely independent, so you’re really dealing with the recurrence $$y_{n+1}=\frac1{2^{y_n}}+\frac12\tag{1}$$ with two different initial values, $2$ and $0$. You need to show that the resulting sequence converges for each of these initial values, and to the same limit for both.

Let $$f(x)=\frac1{2^x}+\frac12\;,$$ and let $g(x)=f(x)-x$. Show that the limit of a convergent sequence satisfying $(1)$ must be a fixed point of $f$ and therefore a zero of $g$. Then show that $g'(x)<0$ for all $x$ and conclude that $g$ has only one zero. By inspection the unique fixed point of $f$ is $x=1$.

Finally, show that if $x<1$, then $x<f\big(f(x)\big)<1$, and if $x>1$, then $x>f\big(f(x)\big)>1$, and put the pieces together to get the desired result.

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To find the limit of the sequence. Assume that $\lim_{n\to \infty} u_n=a$, then

$$\lim_{n\to \infty} u_{n+2}= \lim_{n\to \infty}\frac{1}{2^{u_n}}+\frac{1}{2}$$

$$\implies a = 2^{-a}+1 \,.$$

To solve the above equation you can use the Lambert W function technique. That gives the solution

$$a= {\frac {{W} \left( \frac{\ln(2)}{2} \right) + \ln \left( 2 \right) }{\ln \left( 2 \right) }} = 1.383332348$$

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it should be $a=2^{-a}+1/2$. Also you have not shown that it converges. – DirkGently Sep 6 '15 at 0:00