First slide

Introduction to limits

Question

$I : lim_{n \to \infty} \frac{1 + 3 + 6 + \dots . + \frac{n (n + 1)}{2}}{n^{3}} = \frac{1}{6}$
$II : lim_{n \to \infty} \frac{1.1! + 2.2! + 3.3! + \dots \dots + n . n!}{(n + 1)} = 1$

Moderate

A

Only I is true
B

Only II is true
C

Both I and II are true
D

Neither I nor II is true

Solution

$\lim_{n \to \infty} \frac{\sum \frac{n^{2} + n}{2}}{n^{3}} =$
$= {\lim_{n \to \infty}}^{\frac{1}{2}} \frac{[\frac{n (n + 1) (2 n + 1)}{6} + \frac{n (n + 1)}{2}]}{n^{3}}$
$= {\lim_{x \to \infty}}^{\frac{1}{2}} [\frac{n . n . n (1 + \frac{1}{n}) (2 + \frac{1}{n})}{6 n^{3}} + \frac{n . n (\frac{1}{n} + 1)}{2 n^{3}}]$
$\begin{array}{l} = \frac{1}{2} [\frac{(1 + 0) (2 + 0)}{6} + 0] = \frac{1}{6} \\ 2) lim_{n \to \infty} \frac{(n + 1)! - 1}{(n + 1)!} \\ = lim_{n \to \infty} \frac{(n + 1)!}{(n + 1)!} - \frac{1}{(n + 1)!} \\ = lim_{n \to \infty} 1 - \frac{1}{(n + 1)!} \\ = 1 - 0 = 1 \end{array}$

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