# How to evaluate the limit lim_(n -> oo) sum_(j=1)^n(5j)/n^2?

Aug 29, 2015

#### Explanation:

This is an image of the question. Thank you. i don'y know how to post an image on an question to be clear.

${\lim}_{n \rightarrow \infty} {\sum}_{j = 1}^{n} \frac{5 j}{n} ^ 2$

${\lim}_{n \rightarrow \infty} {\underbrace{{\sum}_{j = 1}^{n} \frac{5 j}{n} ^ 2}}_{n \text{ is a constant here}}$

 = lim_(nrarroo) 5/n^2 underbrace(sum_(j=1)^n j)_((n(n+1))/2

$= {\lim}_{n \rightarrow \infty} \frac{5}{n} ^ 2 \frac{n \left(n + 1\right)}{2}$

$= \frac{5}{2} {\lim}_{n \rightarrow \infty} \frac{n \left(n + 1\right)}{n} ^ 2$

To evaluate ${\lim}_{n \rightarrow \infty} \frac{n \left(n + 1\right)}{n} ^ 2$

Either use:

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

Or use

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

$= {\lim}_{n \rightarrow \infty} \left(1 \cdot \left(1 + \frac{1}{n}\right)\right) = 1 \cdot 1 = 1$

So we finish:

${\lim}_{n \rightarrow \infty} {\sum}_{j = 1}^{n} \frac{5 j}{n} ^ 2 = \frac{5}{2} {\lim}_{n \rightarrow \infty} \frac{n \left(n + 1\right)}{n} ^ 2$

$= \frac{5}{2} \cdot 1 = \frac{5}{2}$