# How do you find the limit of s(n)=64/n^3[(n(n+1)(2n+1))/6] as n->oo?

Dec 10, 2017

${\lim}_{n \to \infty} s \left(n\right) = {\lim}_{n \to \infty} \frac{64}{n} ^ 3 \left[\frac{n \left(n + 1\right) \left(2 n + 1\right)}{6}\right] = 21 \frac{1}{3}$

#### Explanation:

${\lim}_{n \to \infty} s \left(n\right) = {\lim}_{n \to \infty} \frac{64}{n} ^ 3 \left[\frac{n \left(n + 1\right) \left(2 n + 1\right)}{6}\right]$

= ${\lim}_{n \to \infty} \frac{64}{6} \left[\frac{n}{n} \cdot \frac{n + 1}{n} \cdot \frac{2 n + 1}{n}\right]$

= ${\lim}_{n \to \infty} \frac{64}{6} \left[1 \cdot \left(1 + \frac{1}{n}\right) \cdot \left(2 + \frac{1}{n}\right)\right]$

= $\frac{64}{6} \cdot 1 \cdot 1 \cdot 2$

= $\frac{64}{3} = 21 \frac{1}{3}$