Example 3 Find the largest constant $M>0$, such that for any positive integer $n$, there exist positive real sequences $a_{1}, a_{2}, \cdots, a_{n}$ and $b_{1}, b_{2}, \cdots, b_{n}$, satisfying:
$$
\begin{array}{l}
\text { (1) } \sum_{k=1}^{n} b_{k}=1, \\
2 b_{k} \geqslant b_{k-1}+b_{k+1}(k=2,3, \cdots, n-1) ; \\
\text { (2) } a_{k}^{2} \leqslant 1+\sum_{i=1}^{k} a_{i} b_{i}\left(k=1,2, \cdots, n, a_{n}=M\right) .
\end{array}
$$
If $b_{i}$ is given, then $a_{n}$ reaches its maximum when the second condition is an equality, at this time,
$$
a_{k}^{2}-a_{k-1}^{2}=a_{k} b_{k}(k=0,1, \cdots, n-1) \text {, }
$$

where $a_{0}=1$.
$$
\begin{array}{l}
\text { Then } a_{k}=\frac{b_{k}+\sqrt{b_{k}+4 a_{k-1}^{2}}}{2} \\
>\frac{b_{k}+2 a_{k-1}}{2}=a_{k-1}+\frac{b_{k}}{2} .
\end{array}
$$

Summing up, we get
$$
a_{n}>a_{0}+\frac{\sum_{i=1}^{n} b_{i}}{2}=1+\frac{1}{2}=\frac{3}{2},
$$

which means $M=\frac{3}{2}$ satisfies the condition.
Next, we show that this is the best possible.
Since $a_{k}$ is monotonically increasing, we have
$$
a_{1}>1, a_{i}>1 \text {. }
$$

Thus, $a_{k}=\frac{b_{k}+\sqrt{b_{k}^{2}+4 a_{k-1}^{2}}}{2}$
$$
\begin{aligned}
& \frac{b_{k}+\frac{b_{k}^{2}}{4}+2 a_{k-1}}{2} \\
= & a_{k-1}+\frac{b_{k}}{2}+\frac{b_{k}^{2}}{8} .
\end{aligned}
$$

Therefore, $a_{n}<a_{0}+\frac{\sum_{i=1}^{n} b_{i}}{2}+\frac{\sum_{i=1}^{n} b_{i}^{2}}{8}=\frac{3}{2}+\frac{\sum_{i=1}^{n} b_{i}^{2}}{8}$.
We need to show that $\sum_{i=1}^{n} b_{i}^{2}$ can be arbitrarily small.
Using property 3, we know that when $n \rightarrow \infty$,
$$
\sum_{i=1}^{n} b_{i}^{2} \leqslant \frac{2}{n-1}\left(\sum_{i=1}^{n} b_{i}\right)^{2}=\frac{2}{n-1} \rightarrow 0 \text {. }
$$

Therefore, the maximum value of $M$ is $\frac{3}{2}$.