[Example 2.2.2] Let $P(x)$ be an $n$-degree polynomial, and for $k=0,1,2, \cdots, n$, $P(k)=\frac{k}{k+1}$.
Find the value of $P(n+1)$.
To find the value of $P(n+1)$, it is necessary to determine the analytical expression of $P(x)$, which must satisfy the following two conditions:
(1) $P(x)$ is an $n$-degree polynomial;
(2) $P(k)=\frac{k}{k+1}, k=0,1, \cdots, n$.
Directly finding a polynomial $P(x)$ that satisfies both conditions (1) and (2) can be challenging. We can adopt a generalization strategy, which involves first generalizing the analytical expression that satisfies condition (2), i.e., temporarily ignoring whether this expression meets condition (1).
An analytical expression that satisfies condition (2) is easily found, for example,
$$
P(x)=\frac{x}{x+1} .
$$
The advantage of (1) is that it satisfies condition (2); for $k=0,1, \cdots, n$, $P(k)=\frac{k}{k+1}$. The disadvantage is that it does not satisfy condition (1), as it is not a polynomial, let alone an $n$-degree polynomial.

To address this, we further generalize by adding an $n+1$-degree polynomial $Q(x)$ to the numerator of (1). This way, the numerator becomes an $n+1$-degree polynomial, and the denominator remains a first-degree polynomial, bringing us closer to an $n$-degree polynomial, i.e., set
$$
P(x)=\frac{Q(x)+x}{x+1},
$$

where $Q(x)$ is an $n+1$-degree polynomial.
To ensure that (2) satisfies condition (2), $Q(x)$ must be zero when $x=0,1, \cdots, n$.
Thus, we can naturally set $Q(x)$ as
$$
Q(x)=a x(x-1)(x-2) \cdots(x-n) .(a \neq 0)
$$

This is an $n+1$-degree polynomial.
Thus, the problem is reduced to making the polynomial
$$
T(x)=Q(x)+x
$$

divisible by $x+1$.
Therefore, $T(x)$ must have the factor $x+1$, which means
$$
\begin{array}{c}
T(-1)=0 . \\
T(-1)=a(-1)(-2) \cdots(-1-n)+(-1)=0, \\
(-1)^{n+1}(n+1)!\cdot a-1=0, \\
a=\frac{(-1)^{n+1}}{(n+1)!} .
\end{array}
$$

Thus, the required polynomial is
$$
\begin{aligned}
P(x) & =\frac{\frac{(-1)^{n+1}}{(n+1)!} \cdot x(x-1) \cdots(x-n)+x}{x+1} \\
& =\frac{(-1)^{n+1} x(x-1) \cdots(x-n)+(n+1)!x}{(x+1) \cdot(n+1)!},
\end{aligned}
$$

Therefore,
$$
\begin{aligned}
P(n+1) & =\frac{(-1)^{n+1} \cdot(n+1) \cdot n \cdots 2 \cdot 1+(n+1)!(n+1)}{(n+1)!(n+2)} \\
& =\frac{n+1+(-1)^{n+1}}{n+2} .
\end{aligned}
$$