Question

${lim}_{n\to \infty }\frac{1^4+2^4+3^4+...+n^4}{n^5}-{lim}_{n\to \infty }\frac{1^3+2^3+3^3+...+n^3}{n^5}$ is:

• A. $\frac{1}{5}$
• B. $\frac{1}{30}$
• C. $\frac{1}{15}$
• D. $\frac{1}{4}$

Answer 1

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

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

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

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

$=\frac{6}{30}-0=\frac{1}{5}$
Hence, option 'A' is correct.