Question

$\displaystyle\lim_{n\to \infty} \frac{2^K+_4^K+6^K+......+(2n)^K}{n^{K+1}}$ is equal to

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

Answer 1

Answer: (A)

Given limit $= \displaystyle\lim_{n\to\infty} 2^{K} \frac{1^{K}+2^{K}+3^{K}+......+n^{K}}{n^{K+1}}$
$=2^{K} \displaystyle\lim_{n\to\infty} \frac{1}{n} \left(\left(\frac{1}{n}\right)^{K}+\left(\frac{2}{n}\right)^{K}+\left(\frac{3}{4}\right)^{K}+\ldots\ldots\left(\frac{r}{h}\right)^{K}+\ldots\ldots\left(\frac{n}{n}\right)^{K}\right)$
$=2^{K} \lim_{h\to0} \displaystyle\sum_{rh=h}^{nh} h\cdot\left(rh\right)^{K}$
$=2^{K} \int \limits_{0}^{1} x^{K} dx=2^{K} \left|\frac{x^{K+1}}{K+1}\right|_{0}^{1}=\frac{2^{K}}{K+1}$