II. (40 points) Find the largest positive real number $\lambda$ such that for all positive integers $n$ and positive real numbers $a_{i} (i=1,2, \cdots, n)$, we have
$$
1+\sum_{k=1}^{n} \frac{1}{a_{k}^{2}} \geqslant \lambda\left[\sum_{k=1}^{n} \frac{1}{\left(1+\sum_{s=1}^{k} a_{s}\right)^{2}}\right] .
$$
II. Define $S_{k}=\sum_{i=1}^{k} a_{i}+1$, and supplement the definition $S_{0}=1$. First, prove a lemma.
Lemma For any positive integer $k \geqslant 1$, we have $\frac{1}{a_{k}^{2}}+\frac{1}{S_{k-1}^{2}} \geqslant \frac{8}{S_{k}^{2}}$.
Proof Let $S_{k-1}=a_{k} t_{k}$. Then $S_{k}=a_{k}\left(t_{k}+1\right)$.
Thus, equation (1) $\Leftrightarrow\left(1+\frac{1}{t_{k}^{2}}\right)\left(t_{k}+1\right)^{2} \geqslant 8$ $\Leftrightarrow t_{k}^{2}+2 t_{k}+2+\frac{2}{t_{k}}+\frac{1}{t_{k}^{2}} \geqslant 8$.
The last inequality is easily proven by the AM-GM inequality.
Returning to the original problem.
By the lemma, we have $\frac{1}{a_{k}^{2}} \geqslant \frac{8}{S_{k}^{2}}-\frac{1}{S_{k-1}^{2}}$.
Thus, $\sum_{k=1}^{n} \frac{1}{a_{k}^{2}} \geqslant \sum_{k=1}^{n} \frac{7}{S_{k}^{2}}+\frac{1}{S_{n}^{2}}-1$.
Hence, $1+\sum_{k=1}^{n} \frac{1}{a_{k}^{2}} \geqslant 7 \sum_{k=1}^{n} \frac{1}{S_{k}^{2}}$.
Therefore, when $\lambda=7$, the inequality holds.
Next, we show that $\lambda=7$ is the best possible.
Let $a_{k}=2^{k-1}$, and let $n \rightarrow+\infty$, it is easy to see that $\lambda \leqslant 7$.
In conclusion, $\lambda_{\text {max }}=7$.