# How do you use the summation formulas to rewrite the expression Sigma (4j+3)/n^2 as j=1 to n without the summation notation and then use the result to find the sum for n=10, 100, 1000, and 10000?

Nov 13, 2016

${\sum}_{j = 1}^{n} \frac{1}{n} ^ 2 \left(4 j + 3\right) = \frac{2 n + 5}{n}$

#### Explanation:

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

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

We have;

${S}_{n} = \frac{1}{n} ^ 2 \left\{4 \frac{1}{2} n \left(n + 1\right) + 3 n\right\}$
$\therefore {S}_{n} = \frac{1}{n} ^ 2 \left(2 n \left(n + 1\right) + 3 n\right)$
$\therefore {S}_{n} = \frac{1}{n} ^ 2 \left(2 {n}^{2} + 2 n + 3 n\right)$
$\therefore {S}_{n} = \frac{1}{n} ^ 2 \left(2 {n}^{2} + 5 n\right)$
$\therefore {S}_{n} = \frac{1}{n} ^ 2 \left(2 n + 5\right) n$
$\therefore {S}_{n} = \frac{2 n + 5}{n}$

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 ${S}_{n} \rightarrow 2$; This is probably the conclusion of this question]

Now, ${S}_{n} = \frac{2 n + 5}{n}$

$\therefore {S}_{n} = 2 \frac{n}{n} + \frac{5}{n}$
$\therefore {S}_{n} = 2 + \frac{5}{n}$

And so,

${\lim}_{n \rightarrow \infty} {S}_{n} = {\lim}_{n \rightarrow \infty} \left(2 + \frac{5}{n}\right)$
$\therefore {\lim}_{n \rightarrow \infty} {S}_{n} = {\lim}_{n \rightarrow \infty} \left(2\right) + {\lim}_{n \rightarrow \infty} \left(\frac{5}{n}\right)$
$\therefore {\lim}_{n \rightarrow \infty} {S}_{n} = 2 + 5 {\lim}_{n \rightarrow \infty} \left(\frac{1}{n}\right)$
$\therefore {\lim}_{n \rightarrow \infty} {S}_{n} = 2$, $\text{ as } \left({\lim}_{n \rightarrow \infty} \left(\frac{1}{n}\right) = 0\right)$
Which confirms our assumption!