Question

\(\lim_{n \to \infty} \left( \frac{1^2}{1 - n^3} + \frac{2^2}{1 - n^3} + \dots + \frac{n^2}{1 - n^3} \right)\) is equal to

• A. \(\frac{1}{3}\)
• B. \(-\frac{1}{3}\)
• C. \(\frac{1}{6}\)
• D. \(-\frac{1}{6}\)

Hint:

When denominator is negative for large n, careful sign handling is important.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Sum of squares formula: \(\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}\).

Step 2: Detailed Explanation:
Expression = \(\frac{1}{1 - n^3} \sum_{k=1}^n k^2 = \frac{1}{1-n^3} \cdot \frac{n(n+1)(2n+1)}{6}\). = \(\frac{n(n+1)(2n+1)}{6(1-n^3)} = \frac{n^3(1+1/n)(2+1/n)}{6 n^3(1/n^3 - 1)} = \frac{(1+1/n)(2+1/n)}{6(1/n^3 - 1)}\). As \(n \to \infty\), \((1+1/n)(2+1/n) \to 2\), \(1/n^3 \to 0\), so limit = \(\frac{2}{6(-1)} = -\frac{1}{3}\).

Step 3: Final Answer:
\(-\frac{1}{3}\), which corresponds to option (B).