# Limit of the sequence $nx_{n}$ where $x_{n+1} = \log (1 +x_{n})$ [duplicate]

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.

## marked as duplicate by Sil, max_zorn, Lord Shark the Unknown, YuiTo Cheng, LeucippusMay 26 at 7:47

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 $$nx_{n} \rightarrow 2$$