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

Dec 10, 2016

${\sum}_{i = 1}^{n} \frac{2 i + 1}{n} ^ 2 = \frac{n + 2}{n}$

#### Explanation:

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

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

We have;

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

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, ${S}_{n} = \frac{n + 2}{n}$

$\therefore {S}_{n} = 1 + \frac{2}{n}$

And so,

${\lim}_{n \rightarrow \infty} {S}_{n} = {\lim}_{n \rightarrow \infty} \left(1 + \frac{2}{n}\right)$
$\therefore {\lim}_{n \rightarrow \infty} {S}_{n} = {\lim}_{n \rightarrow \infty} \left(1\right) + 2 {\lim}_{n \rightarrow \infty} \left(\frac{1}{n}\right)$
$\therefore {\lim}_{n \rightarrow \infty} {S}_{n} = 1 + 0$
$\therefore {\lim}_{n \rightarrow \infty} {S}_{n} = 1$
Which confirms our assumption!