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

To solve the problem of finding the limit of the given series as \( n \) approaches infinity, we first express it in a better form for calculation.

The given expression is:

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

We can factor the denominator and rewrite each term as:

\(\frac{k^2}{1 - n^3}\) for \( k = 1, 2, \ldots, n \).

This can be summarized as:

\(= \frac{1}{1-n^3} \sum_{k=1}^{n} k^2\)

The sum of squares of the first \( n \) natural numbers is given by:

\(\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\)

Thus, the expression becomes:

\(\frac{1}{1-n^3} \times \frac{n(n+1)(2n+1)}{6}\)

Rewriting it gives:

\(= \frac{n(n+1)(2n+1)}{6 \times (1-n^3)}\)\)

Now, simplifying the limit as \( n \to \infty\):

Divide the numerator and the denominator by \( n^3 \), which gives:

\(= \frac{1}{6} \cdot \frac{n(n+1)(2n+1)}{1-n^3} \approx \frac{\frac{2}{n^2}}{\frac{-1}{n^3}} = \frac{2}{-6} = -\frac{1}{3}\)

Therefore, the limit is:

\(-\frac{1}{3}\)

Hence, the correct answer is \(-\frac{1}{3}\).