Question

lim n→∞ \(\frac{1}{n^3}\) \(\sum_{k=1}^{n} k^{2} =\)

 

• A. x
• B. \(\frac{x}{2}\)
• C. \(\frac{x}{3}\)
• D. \(\frac{x}{4}\)

Answer 1

The Correct Option is C

To solve the problem \(\lim_{n \to \infty} \frac{1}{n^3} \sum_{k=1}^{n} k^2\), we need to evaluate the limit of the given expression as \(n\) approaches infinity.

The expression involves a sum of squares of the first \(n\) natural numbers. The formula for the sum of squares is:

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

Substitute this formula into the original expression:

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

Simplify the expression:

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

As \(n\) approaches infinity, the terms \(n+1\) and \(2n+1\) in the numerator each approach \(n\), allowing us to approximate:

\(\lim_{n \to \infty} \frac{(n+1)(2n+1)}{6n^2} = \lim_{n \to \infty} \frac{2n^2 + 3n + 1}{6n^2}\)

Divide each term by \(n^2\):

\(\lim_{n \to \infty} \left(\frac{2n^2}{6n^2} + \frac{3n}{6n^2} + \frac{1}{6n^2}\right) = \lim_{n \to \infty} \left(\frac{2}{6} + \frac{3}{6n} + \frac{1}{6n^2}\right)\)

As \(n\) approaches infinity, the terms \(\frac{3}{6n}\) and \(\frac{1}{6n^2}\) both approach zero:

\(\lim_{n \to \infty} \left(\frac{2}{6} + 0 + 0\right) = \frac{1}{3}\)

Thus, the value of the limit is:

\(\frac{x}{3}\), given that \(x = 1\).

Therefore, the correct answer is \(\frac{x}{3}\).