What is the limit of this sequence? It is hard.. Let $a_0 > 0$
$a_{n+1} = \ln(a_n + 1)$
Prove that $\displaystyle\lim_{n\to\infty} n\cdot a_n  =  2$
 A: If you're not used to this type of question, here's a process that often works:
1) Prove that here your sequence is decreasing, and bounded:
$\ln(1+x) \leq x \implies a_{n+1} \leq a_n $ , ie $(a_n)$ decreases.
Now you can prove by a trivial induction that $(a_n) \geq 0$. 
$(a_n)$ is lower-bounded, decreasing, hence it converges $\rightarrow L$. 
2) Evaluate L: 
You can take the limit on both side of the recursive relation : $L = \ln(L+1) $
If you study the function $h: x \rightarrow x - \ln(1+x)$ , you'll see that the only possibility to get h(x) = 0 is : x=0. That proves that : L=0. 
3) Pushing it further: $\ln(1+u) = u - \frac{u^2}{2} + o(u^2)$ ; when $u \rightarrow 0$
You get : $ a_{n+1} = a_n - \frac{a_n^2}{2} + o(a_n^2) = a_n(1 - \frac{a_n}{2} + o(a_n)) $
Let:$ v_n = a_n^{-1}$
=> $ v_{n+1} = v_n(1- \frac{a_n}{2} + o(a_n))^{-1} $
$(1-u)^{-1} = 1 + u +o(u)$ ; when $ u \rightarrow 0 $, thus giving:
$ v_{n+1} = v_n(1 +\frac{a_n}{2} + o(a_n)) = v_n + \frac{1}{2} + o(1)  $
=> $ v_{n+1} -v_n = \frac{1}{2} + o(1) $
Here you can say that partial sums of both part are equivalent, since $\sum \frac{1}{2}$ diverges to $+\infty$. 
You get: $ v_{n} - v_0 $ ~ $v_n$ ~ $\frac{1}{2}\sum_{k=0}^{n-1} 1  = \frac{n}{2}$
Finally you get : $ a_n = v_n^{-1} $ ~$ \frac{2}{n}$ ; that is: $ na_n \rightarrow 2$
A: For limit calculation propose another variant that is a combination of Lemma Stolz-Cesaro and rule of L'Hospital:
$$\lim_{n\rightarrow\infty}na_n=\lim_{n\rightarrow\infty}\frac{n}{\frac{1}{a_n}}=\lim_{n\rightarrow\infty}\frac{n+1-n}{\frac{1}{a_{n+1}}-\frac{1}{a_n}}=\lim_{n\rightarrow\infty}\frac{a_n\cdot a_{n+1}}{a_n-a_{n+1}}=\lim_{n\rightarrow\infty}\frac{a_n\cdot \ln (1+a_n)}{a_n-\ln (1+a_n)}=\lim_{x\rightarrow0}\frac{x\cdot \ln (1+x)}{x-\ln (1+x)}=\lim_{x\rightarrow0}\frac{ln(1+x)}{x}\cdot\frac{x^2}{x-\ln (1+x)}=1\cdot\lim_{x\rightarrow0}\frac{2x}{1-\frac{1}{1+x}}=2.$$
