Question about a sequence $a_{n+1}=2^{a_n}-1$ Let $0<t<1$ and define a sequence $(a_n)_{n=1}^\infty$ by:
$$a_{n+1}=2^{a_n}-1 ~~~ , ~~~ a_1=t$$

*

*Prove that $(a_n)_{n=1}^\infty$ is decreasing.

*Prove that $a_{n+1}=2^{a_n}-1$ is convergent and compute its limit.

I have tried to prove 1 in many ways, by induction, and by definition, but I miss something here that doesn't let me continue the proof but I don't know what it is.
I will appreciate some hints and way of thinking for this kind of problem.
Thanks a lot!
 A: It is easy to see that the equation $2^x-1=x$ has two roots $x=0$ and $x=1$. Since $2^x-1-x$ is a convex function (easily verifiable through second order differentiation), these two roots are the only ones. It then immediately follows that $2^x-1<x$ for $x\in[0,1]$. To this reason and since $0<a_n<1$ implies $0<a_{n+1}<1$, we can say that $a_{n+1}<a_n$, which proves the strict decrement of the sequence.
To find the limit, since $a_n\ne 0$ for all $n$ we have
$$
\frac{a_{n+1}}{a_n}=\frac{2^{a_n}-1}{a_n}.
$$
The function $f(x)=\frac{2^x-1}{x}$ is strictly increasing and positive over $(0,1)$ since
$$
f(x)=\sum_{n=0}^\infty \frac{x^n}{(n+1)!}(\ln 2)^{n+1}.
$$
Because $f(1)=1$, for any $t<1$ we have $f(t)<1$, therefore
$$
0<\frac{a_{n+1}}{a_n}=\frac{2^{a_n}-1}{a_n}<\frac{2^{a_1}-1}{a_1}<1,
$$
yielding
$$
0<a_{n+1}<(\frac{2^{a_1}-1}{a_1})a_n<(\frac{2^{a_1}-1}{a_1})^n a_1.
$$
This proves $a_n\to 0$ by squeeze theorem.
A: You are trying to show that $a_{n+1}\leq a_n$, i.e. $$2^{a_n}-1\leq a_n\tag{1}$$  This is not true in general (take $a_n$ to be a large negative number), but it is true when $a_n\in[0,1]$.  Luckily, one can show via induction that $0\leq a_n\leq1$ for all $n$.
To see (1), first note that it holds with equality when $a_n=0$ and when $a_n=1$.  Since each side of (1) is a convex function, the two curves intersect in at most two points.  (Exercise: prove this.)  So either the left-hand side is always less than the right on $[0,1]$, or vice-versa.  Which case is it?
(I originally had the following alternate solution: note that (1) holds with equality at a boundary point (say, $1$).  Then move $a_n$ towards the center: which side of the equation decreases faster?)
(To make this precise, you'll need some Calculus.)
A: To see that $(a_n)$ is decreasing, one could show that stronger statement that $y(x) = 2^x - x - 1$ satisfies $y(x) \leq 0$ for all $x \in [0,1]$. This follows immediately once one observes that $2^x \leq 1 +x$ for $x \in [0,1]$.
To see that $(a_n)$ converges, it suffices to show that $(a_n)$ is bounded below, because bounded monotonic sequences converge. An easy induction argument may be used to show that $a_n \geq 0$ for all $n \geq 1$.
A: First of all, I'll prove that $0<a_i<1\forall i$.
It's easy with induction, the base case is even given. If $0<a_i<1$, $1<2^{a_i}<2\implies 0<2^{a_i}-1<1\implies 0<a_{i+1}<1$, completing the induction.
Now to solve $1$ we just need to prove that $a_{i+1}<a_i$, or $2^{a_i}<a_i+1$. I'll rephrase it:

If $0<x<1$ prove that $2^x<x+1$.

Note that equality holds at $x=1$. As $x$ decreases, the LHS is decreasing exponentially and the RHS is decreasing linearly. If you take the derivative of both sides you can rigorously show which is decreasing faster.
For 2), note that $0<a_{n+1}<a_n<1$. It's bounded from above, and below. You can use the squeeze theorem now to show that $a_n\to 0$.
A: First off, we should note that $x\in(0,1)$ implies that $a(x)=2^x-1\in(0,x)$.
To do so, write $f(x)=x-a(x)$, and note that $f'(x)=1-2^x\ln2$, so that $f'(\ell)=0$ when $$\ell=\log_2\left(\frac{1}{\ln2}\right)\approx 0.5288.$$
Since $f''(x)=-2^x\ln(2)^2<0$, $\ell$ must be a local maximum for $f$. Furthermore, the local minima for $f$ are at $x=0,1$ with $f(0)=f(1)=0$. Therefore $$0<f(x)\le f(\ell)\approx 0.0861,\qquad x\in(0,1),$$
so that
$$0<a(x)<x,\qquad x\in(0,1).$$
This being the case, we have $0<a^n(x)<a^m(x)$ for all $n>m$, where $a^{k+1}(x)=a(a^k(x))$ and $a^0(x)=x$. Therefore $$0<a_n<a_m,\qquad n>m.$$
I haven't gotten around to the computation of the limit yet, but I will update you when I do.
