# How do you find a formula for the sum n terms sum_(i=1)^n (16i)/n^3 and then find the limit as n->oo?

Jan 5, 2017

${\sum}_{i = 1}^{n} \frac{16 i}{n} ^ 3 = \frac{18}{n} ^ 2 \left(n + 1\right)$

${\lim}_{n \rightarrow \infty} {\sum}_{i = 1}^{n} \frac{16 i}{n} ^ 3 = 0$

#### Explanation:

Let ${S}_{n} = {\sum}_{i = 1}^{n} \frac{16 i}{n} ^ 3$
$\therefore {S}_{n} = \frac{16}{n} ^ 3 {\sum}_{i = 1}^{n} i$

And using the standard results:
${\sum}_{r = 1}^{n} r = \frac{1}{2} n \left(n + 1\right)$

We have;

${S}_{n} = \frac{16}{n} ^ 3 \left\{\frac{1}{2} n \left(n + 1\right)\right\}$
$\therefore {S}_{n} = \frac{18}{n} ^ 2 \left(n + 1\right)$

Now we examine the behaviour of ${S}_{n}$ as $n \rightarrow \infty$.
We have;

${S}_{n} = \frac{18}{n} ^ 2 \left(n + 1\right)$
$\therefore {S}_{n} = \frac{18}{n} + \frac{18}{n} ^ 2$
$\therefore {\lim}_{n \rightarrow \infty} {S}_{n} = {\lim}_{n \rightarrow \infty} \left\{\frac{18}{n} + \frac{18}{n} ^ 2\right\}$
$\therefore {\lim}_{n \rightarrow \infty} {S}_{n} = 18 {\lim}_{n \rightarrow \infty} \left(\frac{1}{n}\right) + 18 {\lim}_{n \rightarrow \infty} \left(\frac{1}{n} ^ 2\right)$

And as both $\frac{1}{n} \rightarrow 0$ and $\frac{1}{n} ^ 2 \rightarrow 0$ as $n \rightarrow \infty$, we have;

$\therefore {\lim}_{n \rightarrow \infty} {S}_{n} = 0 + 0$
$\therefore {\lim}_{n \rightarrow \infty} {S}_{n} = 0$