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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.

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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 $nx_{n} \rightarrow 2$