Question

Find the value of $\underset{n\to \infty }{lim}\frac{1+2+3+....n}{n^2}$

• A. 1
• B. 2
• C. $\frac{1}{3}$
• D. $\frac{1}{2}$

Answer 1

The correct option is D

$\frac{1}{2}$

We have, $\underset{n\to \infty }{lim}\frac{1+2+3+....n}{n^2}$

Here, numerator is sum of first n natural number

we know, $\sum n=\frac{n(n+1)}{2}$

So, $\underset{n\to \infty }{lim}\frac{\frac{n(n+1)}{2}}{n^2}$

= $\underset{n\to \infty }{lim}\frac{(n+1)}{2n}$

= $\underset{n\to \infty }{lim}(\frac{1}{2}+\frac{1}{2n})$

$=\frac{1}{2}+\frac{1}{\infty }$

$=\frac{1}{2}+0$

$=\frac{1}{2}$