Question

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

• A. $1$
• B. $1/2$
• C. $1/3$
• D. $0$

Answer 1

Answer: (C)

$1/3$

Solution

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

we know that $1^2+2^2+---n^2=\frac{n(n+1)(2n+1)}{6}$

$=\underset{x\to \infty }{lim}\frac{n(n+1)(2n+1)}{6n^3}$

$=\underset{x\to \infty }{lim}\frac{(n+1)(2n+1)}{6n^2}=\underset{x\to \infty }{lim}\frac{2n^2+3n+1}{6n^2}$

Apply L-Hospital rule twice.

$=\underset{x\to \infty }{lim}\frac{4n+3}{12n}$

$=\frac{4}{12}=\frac{1}{3}$

C is correct.