Is $(y_n)_{n\geq1}$ convergent or not? $(x_n)_{n\geq0}$ is a sequence defined by $x_0>1$ and $x_{n+1}=\log_2(1+x_n)$, $n\geq0$.
The sequence $(y_n)_{n\geq1}$ is defined as follows: $y_n=\frac{1}{2^n}\prod\limits_{i=1}^{n}(1+x_i)$. Is the sequence $(y_n)$ convergent? Thank you very much!
 A: Some suggestions are:


*

*Taking the logarithm of $(y_n)$, you get
$$ \log_2{y_n} = -n + \sum_{i=1}^n \log_2(1 + x_i) = \sum_{i=1}^n (x_{i+1} - 1) $$

*Prove that the sequence $(x_n)$ is monotonically decreasing. As a further hint for this, notice that $x_{n+1} < x_n$ is equivalent to $1 + x_n < 2^{x_n}$. When is this inequality true?

*Investigate how fast $x_n$ approaches its limit.

A: Assume that $f(x)=log_2(1+x)-x$. Thus $f'(x)=\frac{1}{xln2}-1$. Since $x>1$, $f$ is a decreasing function and so the sequence $x_n$ is decreasing. Let $\lim x_n=l$. Then, one can see $2^l=l+1$ and so $l=0$ or $l=1$, but therefore, $l=0$, as $x_n$ is decreasing. So $0<x_i<1$ for all $i\in \mathbb N$.
Now, put $a_n=\sum_{i=i}^n x_i-n$. Since $a_{n+1}-a_n=x_{n+1}+1>0$, $a_n$ is an increasing sequence. Furthermore, we have $x_i<1$ for every $i\in \mathbb N$ and so $a_n=\sum_{i=i}^n x_i-n<0$ and $a_n$ is a bounded sequence. Thus $\sum_{i=i}^n x_i-n$ is convergent.
we have $\prod\limits_{i=1}^{n}2^{x_{n+1}}=\prod\limits_{i=1}^{n}(1+x_i)$ and so $2^{\sum_{i=1}^n  x_i}=\prod\limits_{i=1}^{n}(1+x_i)$. It implies that $2^{\sum_{i=1}^n  x_i-n}=\frac{1}{2^n}\prod\limits_{i=1}^{n}(1+x_i)$. Since $\lim \sum_{i=i}^n x_i-n< \infty $ , $\frac{1}{2^n}\prod\limits_{i=1}^{n}(1+x_i)$ is convergent.
A: You can estimate $\log_2(1+x) \le 1+\frac{x-1}{2\ln(2)}$ and by induction $$x_n \le 1+\frac{x_0-1}{(2\ln(2))^n}$$
With the above answer you get $$\log_2(y_n) \le \sum_{i=1}^n \frac{x_0-1}{(2\ln(2))^{i+1}}$$ which is a geometric series.
