Evaluate the sum sum_(i=1)^n (12i^2(i-1))/n^4 for n=10,100,1000 and 10000?

Jul 17, 2017

$\left.\begin{matrix}n = 10 & \implies \text{sum"=3.168 \\ n=100 & => "sum"=3.019698 \\ n=1000 & => "sum"=3.001996998 \\ n=10000 & => "sum} = 3.00019997\end{matrix}\right.$

Explanation:

Let:

${S}_{n} = {\sum}_{i = 1}^{n} \frac{12 {i}^{2} \left(i - 1\right)}{n} ^ 4$
$\text{ } = \frac{12}{n} ^ 4 {\sum}_{i = 1}^{n} \left\{{i}^{3} - {i}^{2}\right\}$
$\text{ } = \frac{12}{n} ^ 4 \left\{{\sum}_{i = 1}^{n} {i}^{3} - {\sum}_{i = 1}^{n} {i}^{2}\right\}$

And using the standard results:

${\sum}_{r = 1}^{n} {r}^{2} = \frac{1}{6} n \left(n + 1\right) \left(2 n + 1\right)$
${\sum}_{r = 1}^{n} {r}^{3} = \frac{1}{4} {n}^{2} {\left(n + 1\right)}^{2}$

We have;

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

And this has been calculated using Excel for $n = 10 , 100 , 1000 , 10000$

Conclusion: 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]

From the above results it looks as if:

${S}_{n} \approx 3$ as $n \rightarrow \infty$

Let us see if this is actually the case. We can manipulate ${S}_{n}$ as follows;

 S_n = ( (n+1)(3n^2- n - 2)) / n^3 }
$\text{ } = \frac{3 {n}^{3} - {n}^{2} - 2 n + 3 {n}^{2} - n - 2}{n} ^ 3$
$\text{ } = \frac{3 {n}^{3} + 2 {n}^{2} - 3 n - 2}{n} ^ 3$
$\text{ } = 3 + \frac{2}{n} - \frac{3}{n} ^ 2 - \frac{2}{n} ^ 3$

And so,

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

Which confirms our numerical observations!