11. (20 points) Let the function
$$
f(x)=\frac{1+\ln (x+1)}{x} \text {, }
$$
$k$ is a positive integer. When $x>0$, $f(x)>\frac{k}{x+1}$ always holds. Find the maximum value of $k$.

 untranslated part:
$k$ is a positive integer. When $x>0$, $f(x)>\frac{k}{x+1}$ always holds. Find the maximum value of $k$. 

(Note: The last part was already in English, so it remains unchanged.)
11. Since $x>0$, we have
$$
k0)
$$

always holds.
$$
\text{Let } g(x)=\frac{(x+1)[1+\ln (x+1)]}{x}(x>0) \text{.}
$$

Then $g^{\prime}(x)=\frac{x-1-\ln (x+1)}{x^{2}}$.
Let $h(x)=x-1-\ln (x+1)$.
Obviously, $h(2)0$.
If $h\left(x_{0}\right)=0$, then $20$ (since $x>0$), then $h(x)$ is monotonically increasing in $(0,+\infty)$.

When $x \in\left(0, x_{0}\right)$, $h(x)0$, i.e., $g^{\prime}(x) >0$, then $g(x)$ is increasing in the interval $\left(x_{0},+\infty\right)$.
Thus, when $x>0$,
$$
\begin{array}{l}
g(x)_{\text{min}}=g\left(x_{0}\right) \\
=\frac{\left(x_{0}+1\right)\left[1+\ln \left(x_{0}+1\right)\right]}{x_{0}}.
\end{array}
$$

From equation (1) and the above equation, we get $g(x)_{\min }=x_{0}+1$.
Given that $k<g(x)_{\text{min}}$, i.e., $k<x_{0}+1$.
Since $3<x_{0}+1<4$ and $k$ is a positive integer, the maximum value of $k$ is 3.