To solve this problem, we need to determine the value of $P(n+1)$ for the given polynomial $P(x)$ of degree $n$ such that

$$P(k) = \frac{k}{k+1} \quad \text{for } k = 0, 1, 2, \ldots, n.$$

Our goal is to express $P(x)$ as:

$$P(x) = x - \frac{x(x-1)\cdots(x-n)}{n+1}.$$

This assumes $P(x)$ behaves such that $P(k) = \frac{k}{k+1}$ holds for each $k$.

### Step-by-step Process:

1. Define a new polynomial:

$$Q(x) = (x+1)P(x) - x$$

This transformation ensures that:

$$Q(k) = (k+1)P(k) - k = (k+1)\frac{k}{k+1} - k = 0$$

for $k = 0, 1, 2, \ldots, n$.

2. Thus, $Q(x)$ is a polynomial of degree $n+1$ with roots $0, 1, 2, \ldots, n$. Therefore,

$$Q(x) = c(x)(x-1)(x-2)\cdots(x-n)$$

where $c$ is a constant coefficient.

3. Consider the leading term of both sides. Since $P(x)$ is a polynomial of degree $n$, the polynomial:

$$(x+1)P(x) = x^{n+1} + \text{lower degree terms}$$

has a leading term equal to $x^{n+1}$. The polynomial $x$ contributes another $x$ term leading to:

$$Q(x) = x(x-1)(x-2)\cdots(x-n)$$

4. Evaluate $Q(x)$ at $x = n+1$. This determines that:

$$Q(n+1) = c(n+1) \cdot n(n-1)\cdots 1 = (-1)^{n+1}c(n+1)!$$

5. Since $Q(x) = x(x-1)\cdots(x-n)$, $c = 1$.

6. Finally, evaluate $P(n+1)$:

$$Q(n+1) = (n+2)P(n+1) - (n+1)$$

Since $Q(n+1) = (-1)^{n+1}c(n+1)!$,

$$(n+2)P(n+1) - (n+1) = (-1)^{n+1}c(n+1)!$$

Solving for $P(n+1)$, we get:

$$(n+2)P(n+1) = (-1)^{n+1} + (n+1)$$

$$P(n+1) = \frac{(-1)^{n+1} + (n+1)}{n+2}$$

Hence, the value $P(n+1)$ is:

$$\boxed{\frac{(-1)^{n+1} + (n+1)}{n+2}}$$