Question

The value of $\underset{n\to \infty }{lim}\frac{1+2^4+3^4+...+n^4}{n^5}-\underset{n\to \infty }{lim}\frac{1+2^3+3^3+...+n^3}{n^5}$ is

• A. $0$
• B. $\frac{1}{4}$
• C. $\frac{1}{5}$
• D. $\frac{1}{30}$

Answer 1

The correct option is C $\frac{1}{5}$

Using $1^4+2^4+...+n^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$
$=\frac{n(n+1)(6n^3+9n^2+n-1)}{30}$

$\underset{n\to \infty }{lim}\frac{1^4+2^4+3^4+...+n^4}{n^5}-\underset{n\to \infty }{lim}\frac{1^3+2^3+...+n^3}{n^5}=\underset{n\to \infty }{lim}\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30n^5}-\underset{n\to \infty }{lim}\frac{n^2{(n+1)}^2}{4\times n^5}=\frac{6}{30}-0=\frac{1}{5}$

Hence, option 'C' is correct.