# How do you find the limit of s(n)=81/n^4[(n^2(n+1)^2)/4] as n->oo?

Nov 7, 2016

$: {\lim}_{n \rightarrow \infty} s \left(n\right) = \frac{81}{4}$

#### Explanation:

$s \left(n\right) = \frac{81}{n} ^ 4 \left[\frac{{n}^{2} {\left(n + 1\right)}^{2}}{4}\right]$
$\therefore s \left(n\right) = \frac{81}{n} ^ 2 \left[\frac{{\left(n + 1\right)}^{2}}{4}\right]$
$\therefore s \left(n\right) = \frac{81}{n} ^ 2 \left[\frac{{n}^{2} + 2 n + 1}{4}\right]$
$\therefore s \left(n\right) = \frac{81}{4} \left[\frac{{n}^{2} + 2 n + 1}{n} ^ 2\right]$
$\therefore s \left(n\right) = \frac{81}{4} \left[{n}^{2} / {n}^{2} + \frac{2 n}{n} ^ 2 + \frac{1}{n} ^ 2\right]$
$\therefore s \left(n\right) = \frac{81}{4} \left[1 + \frac{2}{n} + \frac{1}{n} ^ 2\right]$

So then,
${\lim}_{n \rightarrow \infty} s \left(n\right) = {\lim}_{n \rightarrow \infty} \frac{81}{4} \left[1 + \frac{2}{n} + \frac{1}{n} ^ 2\right]$
$\therefore {\lim}_{n \rightarrow \infty} s \left(n\right) = \frac{81}{4}$ as both $\frac{2}{n} \rightarrow 0$, and $\frac{1}{n} ^ 2 \rightarrow 0$ as $n \rightarrow \infty$