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

Jan 5, 2017

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

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

#### Explanation:

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

And using the standard results:

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

We have;

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

Now we examine the behaviour of ${S}_{n}$ as $n \rightarrow \infty$.
We have;

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

And as $\frac{1}{n} \rightarrow 0$ as $n \rightarrow \infty$, we have;

$\therefore {\lim}_{n \rightarrow \infty} {S}_{n} = 3$