Show $\lim_{n\to\infty} na_n = 2$

Question

Suppose $a_n$ is real sequence which satisfies $$ a_1>0, \quad a_{n+1}=\ln(a_n+1) \quad (n\geq1)$$

How can I evaluate $$\lim_{n\to\infty}na_n$$?

I just know $$\lim_{n\to\infty} a_n=0$$

But I don't know what should I do for $na_n$.

Answer 1

Write $$\lim\limits_{n\to \infty}na_n=\lim\limits_{n\to \infty}\frac{n}{1/a_n}.$$ Then, by Stolz theorem $$\lim\limits_{n\to \infty}na_n=\lim\limits_{n\to \infty}\frac{(n+1)-n}{1/a_{n+1}-1/a_n}=\lim\limits_{n\to \infty}\frac{a_na_{n+1}}{a_n-a_{n+1}}.$$ Since $a_n\to 0$ we have $a_{n+1}=\ln(a_n+1)\sim a_n-\frac{a_n^2}{2}$. So $$\lim\limits_{n\to \infty}\frac{a_na_{n+1}}{a_n-a_{n+1}}=\lim\limits_{n\to \infty}\frac{a_n(a_n-\frac{a_n^2}{2})}{a_n-(a_n-\frac{a_n^2}{2})}=\lim\limits_{n\to \infty}\frac{a^2_n-\frac{a_n^3}{2}}{\frac{a_n^2}{2}}=\lim\limits_{n\to\infty}(2-a_n)=2.$$