11. (20 points) Given $a_{1}=a>0$, for $n \geqslant 2$, we have $a_{n}=\prod_{i=1}^{n-1}\left(1+a_{i}\right)$. Find the real number $a$, such that for all positive integers $n$, we have $\sum_{k=1}^{n} \frac{1}{1+a_{k}} < c$.
$$
11. From the conditions, we have
$$
\begin{array}{l}
a_{n}=\prod_{i=1}^{n-1}\left(1+a_{i}\right) \\
\Rightarrow a_{n+1}=\prod_{i=1}^{n}\left(1+a_{i}\right) \\
\Rightarrow a_{n+1}=a_{n}\left(1+a_{n}\right) \\
\Rightarrow \frac{1}{1+a_{n}}=\frac{1}{a_{n}}-\frac{1}{a_{n+1}}(n \geqslant 2) \\
\Rightarrow \sum_{k=1}^{n} \frac{1}{1+a_{k}}=\frac{1}{1+a_{1}}+\sum_{k=2}^{n}\left(\frac{1}{a_{k}}-\frac{1}{a_{k+1}}\right) \\
\quad=\frac{1}{1+a_{1}}+\frac{1}{a_{2}}-\frac{1}{a_{n+1}}=\frac{2}{1+a}-\frac{1}{a_{n+1}} .
\end{array}
$$

Since \( a > 0 \), we have
$$
a_{n}=\prod_{i=1}^{n-1}\left(1+a_{i}\right)>0,
$$

and
$$
\begin{array}{l}
a_{n}=\prod_{i=1}^{n-1}\left(1+a_{i}\right)=a_{n-1}\left(1+a_{n-1}\right) \\
>a_{n-1}+1(n>2) .
\end{array}
$$

This indicates that the sequence \(\left\{a_{n}\right\}\) has no upper bound and \(a_{n}>n-1\).
If \(a=4041\), then
$$
\begin{array}{l}
\sum_{k=1}^{n} \frac{1}{1+a_{k}}=\frac{2}{1+a}-\frac{1}{a_{n+1}} \\
=\frac{1}{2021}-\frac{1}{a_{n+1}}0 \text {. }
\end{array}
$$

For any \(n>\frac{1}{\varepsilon}\),
$$
\begin{array}{l}
\text { Hence } \sum_{k=1}^{n} \frac{1}{1+a_{k}}=\frac{2}{1+a}-\frac{1}{a_{n+1}} \\
=\frac{1}{2021}-\frac{1}{a_{n+1}}>\frac{1}{2021}-\frac{1}{n} \\
>\frac{1}{2021}-\varepsilon=c .
\end{array}
$$

In conclusion, \(a=4041\) satisfies the problem's requirements.