How do you use the summation formulas to rewrite the expression Sigma (4i^2(i-1))/n^4 as k=1 to n without the summation notation and then use the result to find the sum for n=10, 100, 1000, and 10000?

Dec 15, 2016

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

Let ${S}_{n} = {\sum}_{i = 1}^{n} \frac{4 {i}^{2} \left(i - 1\right)}{n} ^ 4$
$\therefore {S}_{n} = \frac{4}{n} ^ 4 {\sum}_{i = 1}^{n} \left({i}^{3} - {i}^{2}\right)$
$\therefore {S}_{n} = \frac{4}{n} ^ 4 \left\{{\sum}_{i = 1}^{n} {i}^{3} - {\sum}_{i = 1}^{n} {i}^{2}\right\}$

And using the standard results:
${\sum}_{r = 1}^{n} {r}^{2} = \frac{1}{6} n \left(n + 1\right) \left(2 n + 1\right) \text{ ; } {\sum}_{r = 1}^{n} {r}^{3} = \frac{1}{4} {n}^{2} {\left(n + 1\right)}^{2}$

We have;

$\setminus \setminus \setminus \setminus \setminus {S}_{n} = \frac{4}{n} ^ 4 \left\{\frac{1}{4} {n}^{2} {\left(n + 1\right)}^{2} - \frac{1}{6} n \left(n + 1\right) \left(2 n + 1\right)\right\}$
$\therefore {S}_{n} = \frac{4}{n} ^ 4 \left(\frac{n \left(n + 1\right)}{12}\right) \left\{3 n \left(n + 1\right) - 2 \left(2 n + 1\right)\right\}$
$\therefore {S}_{n} = \frac{\left(n + 1\right)}{3 {n}^{3}} \left\{3 {n}^{2} + 3 n - 4 n - 2\right\}$
$\therefore {S}_{n} = \frac{\left(n + 1\right)}{3 {n}^{3}} \left(3 {n}^{2} - n - 2\right)$

And this has been calculated using Excel for $n = 10 , 100 , 1000 , 10000$ What happens as $n \rightarrow \infty$?

[ NB As an additional task we could possibly conclude that as $n \rightarrow \infty$ then ${S}_{n} \rightarrow 1$; This is probably the conclusion of this question]

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

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

And so,

$\setminus \setminus \setminus \setminus \setminus {\lim}_{n \rightarrow \infty} {S}_{n} = {\lim}_{n \rightarrow \infty} \frac{1}{3} \left(3 + \frac{2}{n} - \frac{3}{n} ^ 2 - \frac{2}{n} ^ 3\right)$
$\therefore {\lim}_{n \rightarrow \infty} {S}_{n} = \frac{1}{3} \left(3 + 0 - 0 - 0\right)$
$\therefore {\lim}_{n \rightarrow \infty} {S}_{n} = 1$

Which confirms our assumption!