Question

$$ \lim_{n \to \infty} \frac{(1^2 - 1)(n-1) + (2^2 - 2)(n-2) + \ldots + ((n-1)^2 - (n-1))}{(1^3 + 2^3 + \ldots + n^3) - (1^2 + 2^2 + \ldots + n^2)} $$ is equal to: 

• A. $\frac{2}{3}$
• B. $\frac{1}{3}$
• C. $\frac{3}{4}$
• D. $\frac{1}{2}$

Answer 1

The Correct Option is B

The problem requires determining the limit of the given expression as \( n \to \infty \):

\[\lim_{n \to \infty} \frac{(1^2 - 1)(n-1) + (2^2 - 2)(n-2) + \ldots + ((n-1)^2 - (n-1))}{(1^3 + 2^3 + \ldots + n^3) - (1^2 + 2^2 + \ldots + n^2)}\]

To solve this, we will simplify the numerator and the denominator separately before evaluating the limit.

1. Numerator analysis:

\[\sum_{k=1}^{n-1} \left( k^2 - k \right)(n-k)\]

• Expansion:

\[\sum_{k=1}^{n-1} (k^2n - k^3 - kn + k^2)\]

• For the limit as \( n \to \infty \), we focus on the highest power terms. The dominant term is approximately \( \frac{n^3}{3} \).

\[(n/3)(n^3) \approx \frac{n^3}{3}\]

1. Denominator analysis:

\[(1^3 + 2^3 + \ldots + n^3) - (1^2 + 2^2 + \ldots + n^2)\]

• Using standard summation formulas:
  • Sum of cubes:

\[\sum_{k=1}^{n} k^3 = \left( \frac{n(n + 1)}{2} \right)^2\]

• Sum of squares:

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

• The leading term of the denominator, focusing on the highest power of \( n \), is approximately \( \frac{n^4}{4} \).

\[\left( \frac{n(n + 1)}{2} \right)^2 \approx \frac{n^4}{4}\]

1. Limit calculation by comparing leading terms:

Numerator's leading term:

\[\frac{n^3}{3}\]

Denominator's leading term:

\[\frac{n^4}{4}\]

1. Evaluate the limit:

\[\lim_{n \to \infty} \frac{\frac{n^3}{3}}{\frac{n^4}{4}} = \lim_{n \to \infty} \frac{4n^3}{3n^4} = \lim_{n \to \infty} \frac{4}{3n}\]

As \( n \to \infty \), this limit approaches \( 0 \).

Correction: The previous calculation had an error. Let's re-evaluate the limit.

The limit is:

\[\lim_{n \to \infty} \frac{\frac{n^3}{3}}{\frac{n^4}{4}} = \lim_{n \to \infty} \frac{4n^3}{3n^4} = \lim_{n \to \infty} \frac{4}{3n}\]

This still leads to 0. There seems to be a misunderstanding in the prompt's provided "solution." Let's re-examine the numerator's leading term more carefully.

The numerator is \( \sum_{k=1}^{n-1} (nk^2 - k^3 - nk + k^2) \). The terms with the highest power of \( n \) come from \( nk^2 \). When summed up to \( n-1 \), this is approximately \( n \sum_{k=1}^{n} k^2 \approx n \left( \frac{n^3}{3} \right) = \frac{n^4}{3} \).

The dominant term of the numerator is \( \frac{n^4}{3} \).

The dominant term of the denominator is \( \frac{n^4}{4} \).

Therefore, the limit is:

\[\lim_{n \to \infty} \frac{\frac{n^4}{3}}{\frac{n^4}{4}} = \lim_{n \to \infty} \frac{4n^4}{3n^4} = \frac{4}{3}\]

The correct answer is:

\[\frac{4}{3}\]

.