Step 1: Understanding the Question:
This is a limit of a sum as \( n \) approaches infinity. We can combine the terms over a common denominator and use the sum of cubes formula. Step 2: Key Formula or Approach:
Sum of first \( n \) cubes \( \sum r^3 = \frac{n^2(n+1)^2}{4} \approx \frac{n^4}{4} \) for large \( n \). Step 3: Detailed Explanation:
The sum is \( \frac{1 + 2^3 + 3^3 + \dots + n^3}{1 - n^4} \).
Numerator \( = \frac{n^2(n+1)^2}{4} \).
Denominator \( = 1 - n^4 \).
Limit:
\[ \lim_{n \to \infty} \frac{n^2(n+1)^2}{4(1 - n^4)} = \lim_{n \to \infty} \frac{n^2 \cdot n^2(1 + 1/n)^2}{4n^4(1/n^4 - 1)} \]
\[ = \lim_{n \to \infty} \frac{n^4(1 + 1/n)^2}{4n^4(1/n^4 - 1)} = \frac{(1 + 0)^2}{4(0 - 1)} = -\frac{1}{4} \]. Step 4: Final Answer:
The limit is \( -1/4 \).