Question

$\underset{n\to \infty }{\lim }$$\frac{1+2+3+...+n}{n^2}$, $n\in N$, is equal to

• A. $0$
• B. $1$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

Answer: (C)

As $\underset{n\to \infty }{\lim }$$\frac{1+2+3+...+n}{n^2}$
$=\underset{n\to \infty }{\lim }$$\frac{n\left(n+1\right)}{2n^2}$
$=\underset{n\to \infty }{\lim }$$\frac{1}{2}\left(1+\frac{1}{n}\right)$
$=\frac{1}{2}$