log sequence limit

Consider the sequence defined as follows, with $x_1 \gt 0$.

$$x_{n+1} = \log (1 + x_n)$$

One can show that $x_n \to 0$ as $n \to \infty$ (try it).

What can you say about the sequence $y_n = nx_n$ as $n \to \infty$? Does the limit exist? If so, what is it?

Note: The $\log$ is the natural logarithm to base $e$.

[Solution]

The \(n^{th}\) root of \(n\)

For the sake of a quick, "pure" math post.

It is a well known fact that \(\lim_{n \to \infty} \sqrt[n]{n} = 1\)

There are multiple proofs of that. One such proof is to use the inequality of Arithmetic and Geometric means

$$ \frac{x_1 + x_2 + \dots + x_n}{n} \ge \sqrt[n]{x_1\cdot x_2\cdots x_n} \quad (x_i \ge 0)$$

Choose \(x_1 = x_2 = \dots = x_{n-2} = 1\) and \(x_{n-1} = x_n = \sqrt{n}\)

Then we get,

$$\frac{n-2 + 2\sqrt{n}}{n} \ge \sqrt[n]{n} \ge 1 $$

And so,

$$ 1 - \frac{2}{n} + \frac{2}{\sqrt{n}} \ge \sqrt[n]{n} \ge 1$$

By the sandwich theorem, the limit is one.