Question

Evaluate: $\underset{n\to \infty }{lim}\frac{(1^2+2^2+3^2+...+n^2)}{[1+2+...+n](1^3+2^3+3^3+...+n^3)}$

• A. $\frac{8}{9}$
• B. $\frac{9}{8}$
• C. $\frac{4}{3}$
• D. $-\frac{4}{3}$

Answer 1

The correct option is A $\frac{8}{9}$

Given: $\underset{n\to \infty }{lim}\frac{(1^2+2^2+3^2+...+n^2)}{[1+2+...+n](1^3+2^3+3^3+...+n^3)}$

The given equation ca be written as

$\underset{n\to \infty }{lim}\frac{{\left[\frac{n(n+1)(2n+1)}{6}\right]}^2}{\frac{n(n+1)}{2}\times \frac{n^2{(n+1)}^2}{4}}$

$=\underset{n\to \infty }{lim}\frac{n^2{(n+1)}^2{(2n+1)}^2\times 2\times 4}{36n(n+1)n^2{(n+1)}^2}$

$=\underset{n\to \infty }{lim}\frac{2{(2n+1)}^2}{9n(n+1)}$

$=\underset{n\to \infty }{lim}\frac{8n^2+8n+2}{9n^2+9n}$

$=\underset{n\to \infty }{lim}\frac{n^2(8+\frac{8}{n}+\frac{2}{n^2})}{n^2(9+\frac{9}{n})}$

$=\underset{n\to \infty }{lim}\frac{8+\frac{8}{n}+\frac{2}{n^2}}{9+\frac{9}{n}}$

$=\frac{8}{9}$