Suppose that $a_{n+1}=\log(1+a_n),~a_1>0$. Then find $\lim_{n \rightarrow \infty} n \cdot a_n$.
I can find $\lim_{n \rightarrow \infty}a_n=0$. But I have no idea how to find $\lim_{n \rightarrow \infty} n\cdot a_n$. (In this problem, $\log$ denotes the natural logarithm.) Could you give me key idea about solving this problem?
Using the Maclaurin series for $\log(1+x)$, the Stolz–Cesàro theorem, and the fact that $\lim_n a_n = 0$,
$$\begin{align*} \lim_n n\cdot a_n &=\lim_n\frac{n}{(a_n)^{-1}}\\ &=\lim_n\frac{(n+1)-n}{(a_{n+1})^{-1}-(a_n)^{-1}}\\ &=\lim_n\frac{a_n a_{n+1}}{a_n-a_{n+1}}\\ &=\lim_n\frac{a_n\log(1+a_n)}{a_n-\log(1+a_n)}\\ &=\lim_n\frac{(a_n)^2 -\frac12(a_n)^3+\frac13(a_n)^4-\cdots}{\frac12(a_n)^2-\frac13(a_n)^3+\frac14(a_n)^4-\cdots}\\ &=2\lim_n\frac{1 -\frac12a_n+\frac13(a_n)^2-\cdots}{1-\frac23a_n+\frac24(a_n)^2-\cdots}\\ &=2. \end{align*}$$
As an alternative to using the Maclaurin series, the limit $$\lim\limits_n\dfrac{a_n\log(1+a_n)}{a_n-\log(1+a_n)}=\lim_{x\to 0}\dfrac{x\log(1+x)}{x-\log(1+x)}=2$$ can be found with two applications of l'Hôpital's rule.
