log sequence limit [Solution]

The problem was:

$x_1 \gt 0$ and $x_{n+1} = \log (1 + x_n)$.

Find $\lim_{n \to \infty} nx_n$.

Note: $x_n$ is monotonically decreasing and bounded below by $0$, and $x_n \to L = 0$ (solution of $L = \log(1+L)$).

Solution

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Did you know there was something like a l'Hopital rule for sequences?

It is called the Stolz–Cesàro theorem. That can be used here, and easily gives us the limit (in fact by using the fact that $x_n \to 0$ and applying the normal l'Hopital rule twice!)

We will not use that though.

Consider the sequence $$y_n = \dfrac{1}{x_{n+1}} = \dfrac{1}{\log(1+x_n)}$$


Now $\log (1 + x_n) = x_n - x_n^2/2 + O(x_n^3)$

Thus $$y_n = \frac{1}{x_n - x_n^2/2 + O(x_n^3)} = \frac{1}{x_n(1 - x_n/2 + O(x_n^2))}$$

$$= \frac{1}{x_n} - \frac{1}{2} + O(x_n)$$

(Using $\dfrac{1}{1-t} = 1 + t + t^2 + \dots$).

Thus $c_n = y_n - y_{n-1} \to \dfrac{1}{2}$ (we can apply Stolz Cesaro here if we want).

Now if $a_n \to L$, then $S_n = \dfrac{a_1 + a_2 + \dots + a_n}{n} \to L$.

Thus applying the above to $c_n$, we get $\dfrac{y_n}{n} \to \dfrac{1}{2}$.

Thus $$n x_n \to 2$$