Deduce the value of the sum from r=1 to infinity of 1/(r(r+1)(r+2)).

Let the general term be $T_r = \frac{1}{r(r+1)(r+2)}$. We express it as:

$$\frac{1}{r(r+1)(r+2)} = \frac{A}{r} + \frac{B}{r+1} + \frac{C}{r+2}$$

Multiplying both sides by $r(r+1)(r+2)$:

$$1 = A(r+1)(r+2) + B r(r+2) + C r(r+1)$$

To find $A$, set $r=0$: $1 = A(1)(2) \implies A = \frac{1}{2}$.
To find $B$, set $r=-1$: $1 = B(-1)(-1+2) \implies 1 = B(-1)(1) \implies B = -1$.
To find $C$, set $r=-2$: $1 = C(-2)(-2+1) \implies 1 = C(-2)(-1) \implies 1 = 2C \implies C = \frac{1}{2}$.
So, the general term is:

$$T_r = \frac{1}{2r} - \frac{1}{r+1} + \frac{1}{2(r+2)}$$

We can rearrange the terms to form a telescoping sum. Let $h(r) = \frac{1}{2r} - \frac{1}{2(r+1)}$.
Then, the general term $T_r$ can be written as:

$$T_r = \left( \frac{1}{2r} - \frac{1}{2(r+1)} \right) - \left( \frac{1}{2(r+1)} - \frac{1}{2(r+2)} \right) = h(r) - h(r+1)$$

The N-th partial sum is given by:

$$S_N = \sum_{r=1}^{N} (h(r) - h(r+1))$$

This is a telescoping sum, which simplifies to:

$$S_N = h(1) - h(N+1)$$

First, calculate $h(1)$:

$$h(1) = \frac{1}{2(1)} - \frac{1}{2(1+1)} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$$

Next, calculate $h(N+1)$:

$$h(N+1) = \frac{1}{2(N+1)} - \frac{1}{2((N+1)+1)} = \frac{1}{2(N+1)} - \frac{1}{2(N+2)}$$

Substitute these into the expression for $S_N$:

$$S_N = \frac{1}{4} - \left( \frac{1}{2(N+1)} - \frac{1}{2(N+2)} \right) = \frac{1}{4} - \frac{1}{2(N+1)} + \frac{1}{2(N+2)}$$

The value of the infinite series is $\lim_{N \to \infty} S_N$:

$$\sum_{r=1}^{\infty} \frac{1}{r(r+1)(r+2)} = \lim_{N \to \infty} \left( \frac{1}{4} - \frac{1}{2(N+1)} + \frac{1}{2(N+2)} \right)$$

As $N \to \infty$, the terms $\frac{1}{2(N+1)}$ and $\frac{1}{2(N+2)}$ both approach 0.

$$\lim_{N \to \infty} S_N = \frac{1}{4} - 0 + 0 = \frac{1}{4}$$

The value of the series is $\frac{1}{4}$.