Suppose $x_{1}>0$, and consider the sequence, $\{x_{n}\}$ defined as follows: $$x_{n+1}=\log(1+x_{n}) \quad n\geq 1 $$ Find the value of $\displaystyle \lim_{n \to \infty} nx_{n}$
I am having trouble solving it. One thing is clear, that since $x_{n}>0$ and $x_{n+1} < x_{n}$, we can have a sequence which converges an $f$ which satisfies $f=\log(1+f)$, so that $f=0$. Any way as to how we can proceed from here.
Using the methods similar to the answer by David Speyer here: Convergence of $\sqrt{n}x_{n}$ where $x_{n+1} = \sin(x_{n})$
(I suggest you read that first)
I believe we get the answer to be $2$.
Use the fact that $$\log (1+x) = x - \frac{x^2}{2} + O(x^3)$$
If $$y_{n} = \frac{1}{x_n}$$
Then we have that
$$y_{n+1} = \frac{1}{x_n(1 - x_{n}/2 + O((x_{n})^2))} = y_{n} + \frac{1}{2} + O(x_{n})$$
This gives rise to (again similar to that answer)
$$\frac{y_{n}}{n} = \frac{1}{2} + \frac{1}{n}O(\sum_{k=1}^{n-1} x_{k})$$
Since $x_{n} \rightarrow 0$ we have that $$\frac{y_{n}}{n} \rightarrow \frac{1}{2}$$
and thus $$nx_{n} \rightarrow 2$$
