First slide

Introduction to limits

Question

$lim_{x \to \infty} \frac{1^{2} \cdot n + 2^{2} \cdot (n - 1) + 3^{2} \cdot (n - 2) + \dots + n^{2} \cdot 1}{1^{3} + 2^{3} \dots + n^{3}},$ is equal to

Moderate

A

$1 / 3$
B

$2 / 3$
C

$1 / 2$
D

$1 / 6$

Solution

$lim_{n \to \infty} \frac{1^{2} \cdot n + 2^{2} \cdot (n - 1) + 3^{2} \cdot (n - 2) \dots + n^{2} \cdot 1}{1^{3} + 2^{3} + \dots + n^{3}}$
$\begin{aligned} = lim_{n \to \infty} \frac{\sum_{r - 1}^{n} r^{2} (n - r + 1))}{\sum_{r = 1}^{n} r^{3}} \\ = lim_{n \to \infty} \frac{\sum_{r = 1}^{n} \{(n + 1) r^{2} - r^{3}\}}{\sum_{r = 1}^{n} r^{3}} \end{aligned}$
$= lim_{n \to \infty} \frac{(n + 1) (\sum_{r = 1}^{n} r^{2}) - (\sum_{r = 1}^{n} r^{3})}{(\sum_{r = 1}^{n} r^{3})}$
$= lim_{n \to \infty} \frac{\{\frac{(n + 1) n (n + 1) (2 n + 1)}{6}\} - \{\frac{n^{2} (n + 1)}^{2}}{4}\}}{{\{\frac{n (n + 1)}{2}\}}^{2}}$
$= lim_{n \to \infty} \{\frac{\frac{2}{6} - \frac{1}{4}}{\frac{1}{4}}\} = \frac{1}{3}$

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