Question

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

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

Answer 1

The correct option is A $\frac{1}{2}$

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

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

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