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.
We have $$0\leq\log(1+x)\leq x\quad \forall x\geq 0$$ so we can deduce that the sequence $(a_n)$ is decreasing and bounded below by $0$ so it's convergent to $\ell$ which satisfy $\log(1+\ell)=\ell$ so $\ell=0$.
We know that (since $a_n\to 0$) $$a_{n+1}=\log(1+a_n)=a_n-\frac{1}{2}a_n^2+o(a_n^2)$$ hence we have $$a_{n+1}^\alpha=a_n^\alpha\left(1-\frac{1}{2}a_n+o(a_n)\right)^\alpha\sim_\infty a_n^\alpha\left(1-\frac{\alpha}{2}a_n\right)\tag{1}$$ and if we choose $\alpha=-1$ and by telescoping we find $$\sum_{k=1}^{n-1}\frac{1}{a_{k+1}}-\frac{1}{a_{k}}=\frac{1}{a_{n}}-\frac{1}{a_{1}}\sim_\infty\frac{n}{2}$$ and therefore we find $$a_n\sim_\infty\frac{2}{n}$$
Now in the equality $(1)$ and always with $\alpha=-1$ we develop with one term of more: $$\frac{1}{a_{n+1}}-\frac{1}{a_{n}}-\frac{1}{2}\sim_\infty\frac{-1}{12} a_n\sim \frac{-1}{6n}$$ so by telescoping we find $$\sum_{k=1}^{n-1}\left(\frac{1}{a_{k+1}}-\frac{1}{a_{k}}-\frac{1}{2}\right)\sim_\infty\frac{1}{a_n}-\frac{n}{2}\sim_\infty \sum_{k=1}^{n-1}\frac{2}{k}\sim_\infty \frac{-1}{6}\log n$$ and finally we find $$\frac{n(na_n-2)}{\log n}\sim_\infty \frac{2}{3}$$
A physisist's approach to the 1st limit (not proof at all): first of all it is clear that $a_n\rightarrow 0$ as $n\rightarrow\infty$. The question is - how? Supose that $a_n=a(n)$ can be modeled by a smooth function of $x$. Then, expanding the left side and right side of the recursion relation into Taylor series to "first order", we find the ODE $$a'(x)=-\frac12a(x)^2$$ with the obvious solution $a(x)=\frac{2}{x}$. So a physicist would conclude that $\lim_{n\rightarrow\infty}na_n=2$.
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*}