To evaluate the infinite sum \( \sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)} \), we employ partial fraction decomposition. First, express the summand as:
\[
\frac{1}{n(n+1)(n+2)} = \frac{A}{n} + \frac{B}{n+1} + \frac{C}{n+2}
\]
Multiplying by the common denominator yields:
\[
1 = A(n+1)(n+2) + Bn(n+2) + Cn(n+1)
\]
Expanding and collecting terms:
\[
1 = (A + B + C)n^2 + (3A + 2B + C)n + 2A
\]
Equating coefficients:
\[
A + B + C = 0
\]
\[
3A + 2B + C = 0
\]
\[
2A = 1
\]
Solving these equations gives \(A = \frac{1}{2}\). Substituting this into the other equations leads to \(B = -\frac{1}{2}\) and \(C = 0\).
The partial fraction decomposition is:
\[
\frac{1}{n(n+1)(n+2)} = \frac{1}{2n} - \frac{1}{n+1} + \frac{1}{2(n+2)}
\]
Substituting this into the series:
\[
\sum_{n=1}^{\infty} \left( \frac{1}{2n} - \frac{1}{n+1} + \frac{1}{2(n+2)} \right)
\]
This is a telescoping series. The sum evaluates to \( \frac{1}{6} \).