`lim_(n to oo) [ 1^2/n^3 + (2^2)/(n^3) + …+ ((n-1)^2)/(n^3)]`

Answer

To solve the limit

\[ \lim_{n \to \infty} \left( \frac{1^2}{n^3} + \frac{2^2}{n^3} + \cdots + \frac{(n-1)^2}{n^3} \right), \]

we can rewrite the expression inside the limit as follows:

1. **Rewrite the sum**:
The expression can be rewritten as:
\[ \lim_{n \to \infty} \frac{1^2 + 2^2 + \cdots + (n-1)^2}{n^3}. \]

2. **Use the formula for the sum of squares**:
The formula for the sum of the squares of the first \(m\) natural numbers is:
\[ \sum_{k=1}^{m} k^2 = \frac{m(m+1)(2m+1)}{6}. \]
For \(m = n-1\), we have:
\[ \sum_{k=1}^{n-1} k^2 = \frac{(n-1)n(2(n-1)+1)}{6} = \frac{(n-1)n(2n-1)}{6}. \]

3. **Substitute the sum into the limit**:
Substitute this result back into the limit:
\[ \lim_{n \to \infty} \frac{\frac{(n-1)n(2n-1)}{6}}{n^3} = \lim_{n \to \infty} \frac{(n-1)n(2n-1)}{6n^3}. \]

4. **Simplify the expression**:
We can simplify the fraction:
\[ = \lim_{n \to \infty} \frac{(n-1)(2n-1)}{6n}. \]
Now, we can expand the numerator:
\[ = \lim_{n \to \infty} \frac{(2n^2 - n - 2n + 1)}{6n} = \lim_{n \to \infty} \frac{2n^2 - 3n + 1}{6n}. \]

5. **Divide each term by \(n\)**:
Now divide each term in the numerator by \(n\):
\[ = \lim_{n \to \infty} \frac{2n - 3 + \frac{1}{n}}{6}. \]

6. **Evaluate the limit**:
As \(n\) approaches infinity, the \(\frac{1}{n}\) term approaches 0:
\[ = \frac{2 \cdot \infty - 3 + 0}{6} = \frac{2}{6} = \frac{1}{3}. \]

Thus, the final answer is:
\[ \boxed{\frac{1}{3}}. \]