Question

$\underset{n\to \infty }{lim}\frac{n{A}_n}{B_n}$ is equal to?

• A. $\frac{3}{5}$
• B. $\frac{5}{4}$
• C. $\frac{5}{3}$
• D. $\frac{1}{5}$

Answer 1

Answer: (B)

$\frac{5}{4}$

$A_n=1^3+2^3+3^3+...+n^3=\frac{n^2{(n+1)}^2}{4}$

$B_n=1^4+2^4+3^4+...+n^4=\frac{n(n+1)(2n+1)}{6}.\frac{3.n^2+3n-1}{5}$

$\underset{n\to \infty }{lim}\frac{n{A}_n}{B_n}$

$=\underset{n\to \infty }{lim}\frac{n^3{(n+1)}^2}{4}\times \frac{6}{n(n+1)(2n+1)}.\frac{5}{3n^2+3n-1}$

$=\frac{15}{2}\underset{n\to \infty }{lim}\frac{n^2{(n+1)}^2}{(n+1)(2n+1)(3n^2+3n-1)}$

$=\frac{15}{2}\times \frac{1}{6}=\frac{5}{4}$

Hence, option 'B' is correct.