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Question
\[\lim_{n \to \infty} \left[ \frac{1^3 + 2^3 + . . . n^3}{\left( n - 1 \right)^4} \right]\]
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Solution
\[\lim_{n \to \infty} \left[ \frac{1^3 + 2^3 + . . . + n^3}{\left( n - 1 \right)^4} \right]\]
\[ = \lim_{n \to \infty} \left[ \frac{\left[ \frac{n\left( n + 1 \right)}{2} \right]^2}{\left( n - 1 \right)^4} \right]\]
\[ = \lim_{n \to \infty} \left[ \frac{n^2 \left( n + 1 \right)^2}{4 \left( n - 1 \right)^4} \right]\]
Dividing the numerator and the denominator by n4:
\[\lim_{n \to \infty} \left[ \frac{\frac{n^2 \left( n + 1 \right)^2}{n^4}}{4\frac{\left( n - 1 \right)^4}{n^4}} \right]\]
\[ = \lim_{n \to \infty} \left[ \frac{\frac{\left( n + 1 \right)^2}{n^2}}{4 \left( \frac{n - 1}{n} \right)^4} \right]\]
\[ = \lim_{n \to \infty} \left[ \frac{\left( 1 + \frac{1}{n} \right)^2}{4 \left( 1 - \frac{1}{n} \right)^4} \right]\]
\[\text{ When } n \to \infty , \text{ then } \frac{1}{n} \to 0 . \]
\[\frac{\left( 1 + 0 \right)^2}{4 \left( 1 - 0 \right)^4}\]
\[ = \frac{1}{4}\]
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