# How do you find a formula for the sum n terms Sigma1/n^3(i-1)^2 and then find the limit as n->oo?

Nov 10, 2016

${\sum}_{i = 1}^{n} \frac{1}{n} ^ 3 {\left(i - 1\right)}^{2} = \frac{1}{6 {n}^{2}} \left(2 n - 1\right) \left(n - 1\right)$

${\lim}_{n \rightarrow \infty} {\sum}_{i = 1}^{n} \frac{1}{n} ^ 3 {\left(i - 1\right)}^{2} = \frac{1}{3}$

#### Explanation:

Let ${S}_{n} = {\sum}_{i = 1}^{n} \frac{1}{n} ^ 3 {\left(i - 1\right)}^{2}$
$\therefore {S}_{n} = \frac{1}{n} ^ 3 {\sum}_{i = 1}^{n} \left({i}^{2} - 2 i + 1\right)$
$\therefore {S}_{n} = \frac{1}{n} ^ 3 \left\{{\sum}_{i = 1}^{n} {i}^{2} - 2 {\sum}_{i = 1}^{n} i + {\sum}_{i = 1}^{n} 1\right\}$

And using the standard results:
${\sum}_{r = 1}^{n} r = \frac{1}{2} n \left(n + 1\right)$ and ${\sum}_{r = 1}^{n} {r}^{2} = \frac{1}{6} n \left(n + 1\right) \left(2 n + 1\right)$

We have;

${S}_{n} = \frac{1}{n} ^ 3 \left\{\frac{1}{6} n \left(n + 1\right) \left(2 n + 1\right) - \left(2\right) \frac{1}{2} n \left(n + 1\right) + n\right\}$
$\therefore {S}_{n} = \frac{n}{6 {n}^{3}} \left\{\left(n + 1\right) \left(2 n + 1\right) - 6 \left(n + 1\right) + 6\right\}$
$\therefore {S}_{n} = \frac{1}{6 {n}^{2}} \left\{2 {n}^{2} + 3 n + 1 - 6 n - 6 + 6\right\}$
$\therefore {S}_{n} = \frac{1}{6 {n}^{2}} \left\{2 {n}^{2} - 3 n + 1\right\}$
$\therefore {S}_{n} = \frac{1}{6 {n}^{2}} \left(2 n - 1\right) \left(n - 1\right)$

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

${S}_{n} = \frac{1}{6 {n}^{2}} \left\{2 {n}^{2} - 3 n + 1\right\}$
$\therefore {S}_{n} = \frac{1}{3} - \frac{1}{2 n} + \frac{1}{6 {n}^{2}}$
$\therefore {\lim}_{n \rightarrow \infty} {S}_{n} = {\lim}_{n \rightarrow \infty} \left\{\frac{1}{3} - \frac{1}{2 n} + \frac{1}{6 {n}^{2}}\right\}$
$\therefore {\lim}_{n \rightarrow \infty} {S}_{n} = {\lim}_{n \rightarrow \infty} \left(\frac{1}{3}\right) - {\lim}_{n \rightarrow \infty} \left(\frac{1}{2 n}\right) + {\lim}_{n \rightarrow \infty} \left(\frac{1}{6 {n}^{2}}\right)$

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

$\therefore {\lim}_{n \rightarrow \infty} {S}_{n} = \frac{1}{3} - 0 + 0$
$\therefore {\lim}_{n \rightarrow \infty} {S}_{n} = \frac{1}{3}$