Question

Evaluate : $\underset{n\to \infty }{lim}\frac{1+2+3+4+\dots .n}{n^2}$

• A. $0$
• B. $\frac{1}{2}$
• C. $\frac{1}{10}$
• D. $\frac{1}{12}$

Answer 1

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

Using, the sum of the first n consecutive natural numbers $=\frac{\left(n\right)(n+1)}{2}$

The expression reduces to,

${lim}_{n\to \infty }\frac{\frac{\left(n\right)(n+1)}{2}}{n^2}$ $={lim}_{n\to \infty }\frac{n^2(1+\frac{1}{n})}{2n^2}=\frac{1+\frac{1}{n}}{2}=\frac{1}{2}$ $[\because n\to \infty \Rightarrow \frac{1}{n}\to 0]$

Hence, option 'B' is correct.