Given a sequence $(u_n)$ defined as \begin{align*} &u_0 = 0\\ &u_{n+1} = u_n + \frac{1}{f(u_n)}, \end{align*} where $f(x) = e^x$. Show that $$u_n = \ln(n) + \frac{\ln(n)}{2n} + o(\frac{\ln(n)}{n}).$$
Could you give me any hint for this problem since I do not know how to start with this.
It's easy to prove that the sequence $(u_n)$ diverges: $u_n \xrightarrow{n\to +\infty}+\infty$
Let us denote $x_n = e^{u_n}$, then $$x_{n+1} = x_n\cdot e^{\frac{1}{x_n}} \tag{1}$$
From $(1)$: $$ x_{n+1} - x_n =x_n \left(e^{\frac{1}{x_n}} - 1 \right) =x_n \left(1 + \frac{1}{x_n}+ \mathcal{O}\left( \frac{1}{x_n^2} \right) -1 \right) = 1 + \mathcal{O}\left( \frac{1}{x_n} \right)$$
we apply the Stolz–Cesàro theorem: $$\frac{x_n}{n} = 1 + \mathcal{O}\left( \frac{1}{x_n} \right)$$
$$ \iff \color{red}{x_n= n + \mathcal{O}\left( \frac{n}{x_n} \right)} \tag{2}$$
We try to develop one more asymtotic term in $(1)$: $$\begin{align} x_{n+1} - x_n &=x_n \left(e^{\frac{1}{x_n}} - 1 \right) \\ &=x_n \left(1 + \frac{1}{x_n}+ \frac{1}{2x_n^2} +\mathcal{O}\left( \frac{1}{x_n^3} \right)-1 \right) \\ &= 1 + \frac{1}{2x_n} +\mathcal{O}\left( \frac{1}{x_n^2} \right) \tag{3} \end{align}$$
From $(2)$ and $(3)$, we deduce that $$x_{n+1} - x_n = 1+ \frac{1}{2n} + \mathcal{O}\left( \frac{1}{n^2} \right) $$ then $$\begin{align} x_n &= x_1 + \sum_{i=1}^{n-1} (x_{i+1}-x_i) \\ &=x_1 +(n-1)+\frac{1}{2}\underbrace{\sum_{i=1}^{n-1} \frac{1}{i}}_{\sim \ln(n)}+ \underbrace{\mathcal{O}\left(\sum_{i=1}^{n-1} \frac{1}{i^2} \right)}_{\text{converge, }=\mathcal{o}\left(\ln(n) \right) }\\ \color{red}{x_n} &\color{red}{= n+\frac{1}{2}\ln(n) + \mathcal{o}\left(\ln(n) \right)} \tag{4} \end{align}$$
From $(4)$, it's easy to prove that
$$u_n = \ln (x_n) = \ln \left(n+\frac{1}{2}\ln(n) + \mathcal{o}\left(\ln(n) \right) \right) = \ln(n) + \frac{\ln(n)}{2n} + o\left(\frac{\ln(n)}{n}\right)$$
Q.E.D
