Example 12 Find the largest constant $M>0$, such that for any positive integer $n$, there exist positive real number sequences $a_{1}, a_{2}, \cdots, a_{n}$ and $b_{1}, b_{2}, \cdots, b_{n}$, satisfying:
(1) $\sum_{k=1}^{n} b_{k}=1, 2 b_{k} \geqslant b_{k-1}+b_{k+1}, k=2,3, \cdots, n-1$;
(2) $a_{k}^{2} \leqslant 1+\sum_{i=1}^{k} a_{i} b_{t}, k=1,2, \cdots, n$, and $a_{n}=M$.
Prove a lemma first: $\max _{1 \leqslant k \leqslant n} a_{k} < 2$ and $\sum_{k=1}^{n} b_{k}=1$, then for any $1 \leqslant m \leqslant n$, we have
$$
\sum_{k=1}^{m} b_{k} \geqslant \frac{m}{2}, \quad \sum_{k=m}^{n} b_{k} \geqslant \frac{n-m+1}{2}
$$

Since $b_{k}>0$, we get from the above
$$
b_{k}>\left\{\begin{array}{l}
\frac{k-1}{m-1} b_{m}, 1 \leqslant k \leqslant m \\
\frac{n-k}{n-m} b_{m}, m \leqslant k \leqslant n
\end{array}\right.
$$

Therefore,
$$
\begin{array}{c}
1=\sum_{k=1}^{n} b_{k}=\sum_{k=1}^{m} b_{k}+\sum_{k=m+1}^{n} b_{k}>\frac{m}{2} b_{m} \\
+\frac{n-m-1}{2} b_{m}=\frac{n-1}{2} b_{m} \\
\quad b_{m}<\frac{2}{n-1}
\end{array}
$$

The lemma is proved.
$$
\begin{array}{c}
\text { Let } f_{0}=1, f_{k}=1+\sum_{i=1}^{k} a_{i} b_{i}, k=1,2, \cdots, n \text {, then } \\
f_{k}-f_{k-1}=a_{k} b_{k} .
\end{array}
$$

By condition (2), we know $a_{k}^{2} \leqslant f_{k}$, i.e., $a_{k} \leqslant \sqrt{f_{k}}(k=1, \cdots, n)$. Also, the lemma shows $a_{k}<2(k=$ $1,2, \cdots, n)$, so from the above we get
$$
f_{k}-f_{k-1}<b_{k} \sqrt{f_{k}} \text {, and } f_{k}-f_{k-1}<2 b_{k} \text {. }
$$

Now we have
$$
\begin{aligned}
& \sqrt{f_{k}}-\sqrt{f_{k-1}}<b_{k} \cdot \frac{\sqrt{f_{k}}}{\sqrt{f_{k}}+\sqrt{f_{k-1}}} \\
= & b_{k}\left(\frac{1}{2}+\frac{f_{k}-f_{k-1}}{2\left(\sqrt{f_{k}}+\sqrt{f_{k-1}}\right)^{2}}\right)<b_{k}\left(\frac{1}{2}+\frac{2 b_{k}}{2\left(\sqrt{f_{k}}+\sqrt{f_{k-1}}\right)^{2}}\right) \\
< & b_{k}\left(\frac{1}{2}+\frac{b_{k}}{4}\right)<b_{k}\left(\frac{1}{2}+\frac{1}{2(n-1)}\right)
\end{aligned}
$$

Summing over $k$ from 1 to $n$, we get
$$
\begin{aligned}
M & =a_{n} \leqslant \sqrt{f_{n}}<\sqrt{f_{0}}+\sum_{k=1}^{n} b_{k} \cdot\left(\frac{1}{2}+\frac{1}{2(n-1)}\right) \\
& =\frac{3}{2}+\frac{1}{2(n-1)}
\end{aligned}
$$

By the arbitrariness of $n$, we get $M_{\max } \leqslant \frac{3}{2}$.
$M=\frac{3}{2}$ can be achieved. For example:
$$
a_{k}=1+\frac{k}{2 n}, b_{k}=\frac{1}{n}, k=1,2, \cdots, n
$$

Then $a_{k}^{2}=\left(1+\frac{k}{2 n}\right)^{2} \leqslant 1+\sum_{i=1}^{k} \frac{1}{n}\left(1+\frac{i}{2 n}\right)$ holds.
In summary, the maximum value of $M$ is $\frac{3}{2}$.