Question

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

• A. $5$
• B. $2$
• C. $1$
• D. $0$

Answer 1

Answer: (D)

$0$

$1+2+3+....+n=\frac{n(n+1)}{2}$

$1^2+2^2+3^2+....+n^2=\frac{n(n+1)(2n+1)}{6}$

$\underset{n\to \infty }{lim}\frac{\sum n}{\sum n^2}=\underset{n\to \infty }{lim}\frac{n(n+1)}{2}\times \frac{6}{n(n+1)(2n+1)}$

$=\underset{n\to \infty }{lim}\frac{3}{2n+1}=\frac{3}{\infty }=0$