Example 5 Find natural numbers $a$, $b$, $c$, such that for any $n \in$ N, $n>2$, we have
$$
\begin{array}{l}
b-\frac{c}{(n-2)!} \\
<\frac{2^{3}-a}{2!}+\frac{3^{3}-a}{3!}+\cdots+\frac{n^{3}-a}{n!}<b .
\end{array}
$$
(1996, World Inter-Cities Mathematical Competition)
Solution: Let $b_{n}=b-\frac{2^{3}-a}{2!}-\frac{3^{3}-a}{3!}-\cdots$ $-\frac{n^{3}-a}{n!}=\frac{P(n)}{n!}$
where $P(n)$ is a polynomial in $n$.
Obviously, the degree of $P(n)$ cannot be greater than 2.
Let $P(n)=k n^{2}+l n+m$.
From $b_{n+1}-b_{n}=-\frac{(n+1)^{3}-a}{(n+1)!}$
$$
=\frac{P(n+1)}{(n+1)!}-\frac{P(n)}{n!}
$$

and by the method of undetermined coefficients, we get $k=1, l=3, m=5, a=5$.
$$
\begin{array}{l} 
\text { Thus, } b= \frac{2^{3}-5}{2!}+\frac{3^{3}-5}{3!}+\cdots+\frac{n^{3}-5}{n!}+ \\
\frac{n^{3}+3 n+5}{n!} .(n>2)
\end{array}
$$

Taking $n+1$ and $n$ in the above formula, and subtracting the two equations, we know that $b$ is a constant. Let $n=3$, we get $b=9$.
$$
\begin{array}{l}
\text { From } b_{n}(n-2)!\cdot b_{n} \\
=(n-2)!\cdot \frac{n^{2}+3 n+5}{n!}=1+\frac{4 n+5}{n^{2}-n} .
\end{array}
$$

Noting that $\frac{4 n+5}{n^{2}-n}$ is a decreasing function of $n$ when $n>2$, thus
$$
c>1+\frac{4 \times 3+5}{3^{2}-3}=\frac{23}{6} \text {. }
$$

We can take $c=4$. Then
$$
a=5, b=9, c_{\text {min }}=4 \text {. }
$$

Comment: (1) Using a similar method, we can consider more general cases: Determine natural numbers $a, b, c$, such that for any $n, k \in \mathbf{N}$, $n>k-1$, we have
$$
\begin{array}{l}
b-\frac{c}{(n-k+1)!}2),
$$
$$
\begin{array}{l}
2^{\circ} \frac{2^{3}-5}{2!}+\frac{3^{3}-5}{3!}+\cdots+\frac{n^{3}-5}{n!}+ \\
\frac{n^{2}+3 n+5}{n!}=9(n>2) \text {, } \\
3^{\circ} \frac{2^{5}-52}{2!}+\frac{3^{5}-52}{3!}+\cdots+\frac{n^{5}-52}{n!}+ \\
\frac{n^{4}+5 n^{3}+14 n^{2}+31 n+52}{n!}=103 \\
\end{array}
$$

These results provide rich material for mathematics competitions.