10. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, a_{n+1}=\frac{a_{n}}{2}+2^{n-1} \cdot a_{n}^{2}\left(n \in \mathbf{N}^{*}\right)$. Let $S_{n}=$ $\sum_{i=1}^{n} \frac{1}{2^{n} a_{n}+1}$, find the value of $\lim _{n \rightarrow \infty} S_{n}$.
10. Let $b_{n}=\frac{1}{2^{n} a_{n}+1}$, then $S_{n}=\sum_{i=1}^{n} b_{i}$.

For $a_{n+1}=\frac{a_{n}}{2}+2^{n-1} \cdot a_{n}^{2}$, multiplying both sides by $2^{n+1}$, we get
$$
2^{n+1} a_{n+1}=2^{n} a_{n}+\left(2^{n} a_{n}\right)^{2} .
$$

Let $c_{n}=2^{n} a_{n}$, then $c_{1}=1, c_{n+1}=c_{n}+c_{n}^{2}$, and $b_{n}=\frac{1}{c_{n}+1}$.
From $c_{n+1}=c_{n}+c_{n}^{2}$, we get $c_{n+1}>c_{n}^{2}$, so $c_{n}>c_{1}^{2^{n-1}}=2^{2^{n-1}}$, hence $\lim _{n \rightarrow \infty} c_{n}=+\infty$.
Since $c_{n+1}=c_{n}+c_{n}^{2}=c_{n}\left(c_{n}+1\right)$, we have
$$
\frac{1}{c_{n+1}}=\frac{1}{c_{n}\left(c_{n}+1\right)}=\frac{1}{c_{n}}-\frac{1}{c_{n}+1} \text {, }
$$

Therefore,
$$
b_{n}=\frac{1}{c_{n}+1}=\frac{1}{c_{n}}-\frac{1}{c_{n+1}} .
$$

Thus,
$$
S_{n}=\sum_{i=1}^{n} b_{i}=\sum_{i=1}^{n}\left(\frac{1}{c_{i}}-\frac{1}{c_{i+1}}\right)=\frac{1}{c_{1}}-\frac{1}{c_{n+1}} .
$$

Therefore,
$$
\lim _{n \rightarrow \infty} S_{n}=\frac{1}{c_{1}}=\frac{1}{2} .
$$