How to find the limit of these sequences? Let $\{a_n\}$ be a real-valued sequence such that $a_1 \geq 0$ and $$a_{n+1}=\ln(a_{n}+1)$$ for all $n\ge1$. How can we find the following limits?
$$\lim_{n\to \infty}na_n=?,$$ $$\lim_{n\to \infty}\frac{n(na_n-2)}{\ln n}=?$$
Thanks in advance.

If you use Stolz–Cesàro theorem, it is much easier to get the limit. Note that $\lim_{n\to\infty}a_n=0$. By using Stolz–Cesàro theorem, 
\begin{eqnarray*}
\lim_{n\to\infty}na_n&=&\lim_{n\to\infty}\frac{n}{\frac{1}{a_n}}=\lim_{n\to\infty}\frac{1}{\frac{1}{a_{n+1}}-\frac{1}{a_n}}=\lim_{n\to\infty}\frac{a_na_{n+1}}{a_n-a_{n+1}}\\
&=&\lim_{n\to\infty}\frac{a_n\ln(a_n+1)}{a_n-\ln(a_n+1)}=\lim_{x\to 0}\frac{x\ln(x+1)}{x-\ln(x+1)}\\
&=&\lim_{x\to 0}\frac{x(x-\frac{1}{2}x^2+O(x^3))}{x-(x-\frac{1}{2}x^2+O(x^3))}\\
&=&2. 
\end{eqnarray*}
Here $\ln(x+1)=x-\frac{1}{2}x^2+O(x^3)$. For the second part, we first note that
$$ a_n=\frac{2}{n}+o(1), \frac{1}{\ln(1+x)}=\frac{1}{x}+\frac{1}{2}-\frac{x}{12}+O(x^2) $$
and hence
\begin{eqnarray*} 
\frac{1}{a_{n+1}}&=&\frac{1}{\ln(1+a_n)}=\frac{1}{a_n}+\frac{1}{2}-\frac{1}{12}a_n+O(a_n^2)=\frac{1}{a_n}+\frac{1}{2}-\frac{1}{6n}+O(\frac{1}{n^2}).
\end{eqnarray*}
So by using Stolz–Cesàro theorem
\begin{eqnarray*} 
\lim_{n\to\infty}\frac{n(na_n-2)}{\ln n}&=&\lim_{n\to\infty}\frac{na_n(na_n-2)}{a_n\ln n}\\
&=&2\lim_{n\to\infty}\frac{na_n-2}{a_n\ln n}=4\lim_{n\to\infty}\frac{\frac{n}{2}-\frac{1}{a_n}}{\ln n}\\
&=&4\lim_{n\to\infty}\frac{(\frac{n+1}{2}-\frac{1}{a_{n+1}})-(\frac{n}{2}-\frac{1}{a_n})}{\ln (n+1)-\ln n}\\
&=&4\lim_{n\to\infty}\frac{\frac{1}{2}-\frac{1}{a_{n+1}}+\frac{1}{a_n}}{\ln (n+1)-\ln n}\\
&=&4\lim_{n\to\infty}\frac{\frac{1}{6n}+O(\frac{1}{n^2})}{\ln (n+1)-\ln n}\\
&=&\frac{2}{3}\lim_{n\to\infty}\frac{1}{n\ln(1+\frac{1}{n})}\\
&=&\frac{2}{3}.
\end{eqnarray*}
