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

Nov 15, 2016

${\sum}_{i = 1}^{n} {\left(1 + \frac{2 i}{n}\right)}^{3} \left(\frac{2}{n}\right) = \frac{2}{n} ^ 2 \left(10 {n}^{2} + 13 n + 4\right)$
${\lim}_{n \rightarrow \infty} {\sum}_{i = 1}^{n} {\left(1 + \frac{2 i}{n}\right)}^{3} \left(\frac{2}{n}\right) = 20$

#### Explanation:

We will need the standard results:
${\sum}_{r = 1}^{n} r = \frac{1}{2} n \left(n + 1\right)$
${\sum}_{r = 1}^{n} {r}^{2} = \frac{1}{6} n \left(n + 1\right) \left(2 n + 1\right)$
${\sum}_{r = 1}^{n} {r}^{2} = \frac{1}{4} {n}^{2} {\left(n + 1\right)}^{2} = {\left({\sum}_{r = 1}^{n} r\right)}^{2}$

Let

${S}_{n} = {\sum}_{i = 1}^{n} {\left(1 + \frac{2 i}{n}\right)}^{3} \left(\frac{2}{n}\right)$
$\therefore {S}_{n} = \frac{2}{n} {\sum}_{i = 1}^{n} {\left(\frac{n + 2 i}{n}\right)}^{3}$
$\therefore {S}_{n} = \frac{2}{n} {\sum}_{i = 1}^{n} \frac{1}{n} ^ 3 {\left(n + 2 i\right)}^{3}$
$\therefore {S}_{n} = \frac{2}{n} ^ 4 {\sum}_{i = 1}^{n} {\left(n + 2 i\right)}^{3}$
$\therefore {S}_{n} = \frac{2}{n} ^ 4 {\sum}_{i = 1}^{n} \left({n}^{3} + 3 {n}^{2} \left(2 i\right) + 3 n {\left(2 i\right)}^{2} + {\left(2 i\right)}^{3}\right)$
$\therefore {S}_{n} = \frac{2}{n} ^ 4 {\sum}_{i = 1}^{n} \left({n}^{3} + 6 {n}^{2} i + 12 n {i}^{2} + 8 {i}^{3}\right)$
$\therefore {S}_{n} = \frac{2}{n} ^ 4 \left\{{\sum}_{i = 1}^{n} {n}^{3} + {\sum}_{i = 1}^{n} 6 {n}^{2} i + {\sum}_{i = 1}^{n} 12 n {i}^{2} + {\sum}_{i = 1}^{n} 8 {i}^{3}\right\}$
$\therefore {S}_{n} = \frac{2}{n} ^ 4 \left\{{n}^{3} {\sum}_{i = 1}^{n} 1 + 6 {n}^{2} {\sum}_{i = 1}^{n} i + 12 n {\sum}_{i = 1}^{n} {i}^{2} + 8 {\sum}_{i = 1}^{n} {i}^{3}\right\}$
$\therefore {S}_{n} = \frac{2}{n} ^ 4 \left\{{n}^{3} n + 6 {n}^{2} \frac{1}{2} n \left(n + 1\right) + 12 n \frac{1}{6} n \left(n + 1\right) \left(2 n + 1\right) + 8 \frac{1}{4} {n}^{2} {\left(n + 1\right)}^{2}\right\}$
$\therefore {S}_{n} = \frac{2}{n} ^ 4 \left\{{n}^{4} + 3 {n}^{3} \left(n + 1\right) + 2 {n}^{2} \left(n + 1\right) \left(2 n + 1\right) + 2 {n}^{2} {\left(n + 1\right)}^{2}\right\}$
$\therefore {S}_{n} = \frac{2}{n} ^ 4 {n}^{2} \left\{{n}^{2} + 3 n \left(n + 1\right) + 2 \left(n + 1\right) \left(2 n + 1\right) + 2 {\left(n + 1\right)}^{2}\right\}$
$\therefore {S}_{n} = \frac{2}{n} ^ 2 \left\{{n}^{2} + 3 {n}^{2} + 3 n + 2 \left(2 {n}^{2} + 3 n + 1\right) + 2 \left({n}^{2} + 2 n + 1\right)\right\}$
$\therefore {S}_{n} = \frac{2}{n} ^ 2 \left\{{n}^{2} + 3 {n}^{2} + 3 n + 4 {n}^{2} + 6 n + 2 + 2 {n}^{2} + 4 n + 2\right\}$
$\therefore {S}_{n} = \frac{2}{n} ^ 2 \left(10 {n}^{2} + 13 n + 4\right)$

Hence,

${\sum}_{i = 1}^{n} {\left(1 + \frac{2 i}{n}\right)}^{3} \left(\frac{2}{n}\right) = \frac{2}{n} ^ 2 \left(10 {n}^{2} + 13 n + 4\right)$

And so;

${\lim}_{n \rightarrow \infty} {S}_{n} = \frac{2}{n} ^ 2 \left(10 {n}^{2} + 13 n + 4\right)$
${\lim}_{n \rightarrow \infty} {S}_{n} = 20 + \frac{26}{n} + \frac{8}{n} ^ 2$
${\lim}_{n \rightarrow \infty} {S}_{n} = 20$ as both $\frac{1}{n} \rightarrow 0$, and $\frac{1}{n} ^ 2 \rightarrow 0$ as $n \rightarrow \infty$

Hence,

${\lim}_{n \rightarrow \infty} {\sum}_{i = 1}^{n} {\left(1 + \frac{2 i}{n}\right)}^{3} \left(\frac{2}{n}\right) = 20$