\( \lim _ { n \rightarrow \infty } ( \frac { 1 } { n ^ { 2 } } + \frac { 2 } { n ^ { 2 } } + \frac { 3 } { n ^ { 2 } } + \cdots + \frac { n - 1 } { n ^ { 2 } } ) = \frac { 1 } { 2 } \)

To compute the limit, let's first express the given series explicitly. We can rewrite the sum as: \[ \lim_{n \to \infty} \left( \frac{1 + 2 + 3 + \cdots + (n-1)}{n^2} \right) \] The sum \(1 + 2 + 3 + \cdots + (n-1)\) can be simplified using the formula for the sum of the first \(m\) integers: \[ \frac{(n-1)n}{2} \] Thus, the limit transforms to: \[ \lim_{n \to \infty} \frac{\frac{(n-1)n}{2}}{n^2} = \lim_{n \to \infty} \frac{(n-1)}{2n} = \lim_{n \to \infty} \frac{1 - \frac{1}{n}}{2} \] As \(n\) approaches infinity, \(\frac{1}{n}\) approaches \(0\). Therefore, the limit simplifies to: \[ \frac{1}{2} \] So, indeed we have: \[ \lim _ { n \rightarrow \infty } \left( \frac { 1 } { n ^ { 2 } } + \frac { 2 } { n ^ { 2 } } + \frac { 3 } { n ^ { 2 } } + \cdots + \frac { n - 1 } { n ^ { 2 } } \right) = \frac { 1 } { 2 } \]