Question

$\underset{n\to \infty }{lim}\frac{2^k+4^k+6^k+...+{\left(2n\right)}^k}{n^{k+1}},k\ne 1,$ is equal to

• A. $2^k$
• B. $\frac{2^k}{k+1}$
• C. $\frac{1}{k+1}$
• D. none of these

Answer 1

Answer: (B)

$\frac{2^k}{k+1}$

${lim}_{n\to \infty }\frac{2^k(1^k+2^k+3^k+...+n^k)}{n{n}^k}$

$={lim}_{n\to \infty }\frac{2^k}{n}({\left(\frac{1}{n}\right)}^k+{\left(\frac{2}{n}\right)}^k+....+{\left(\frac{n}{n}\right)}^k)$

$=2^k{lim}_{n\to \infty }\frac{1}{n}{\sum }_{r=1}^n{\left(\frac{r}{n}\right)}^k({lim}_{n\to \infty }\frac{1}{n}{\sum }_{r=1}^{n}f\left(\frac{r}{n}\right)={\int }_0^{1}f\left(x\right)dx)$

$=2^k\times {\int }_0^1{x}^{k}dx$

$=2^k\times \frac{{\left[x^{k+1}\right]}_0^1}{k+1}=\frac{2^k}{k+1}(1-0)=\frac{2^k}{k+1}$