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Question: \[\lim_{n \rightarrow \infty}\frac{1^{p} + 2^{p} + 3^{p} + ..... + n^{p}}{n^{p + 1}} =\]...

lim⁡n→∞1p+2p+3p+.....+npnp+1=\lim_{n \rightarrow \infty}\frac{1^{p} + 2^{p} + 3^{p} + ..... + n^{p}}{n^{p + 1}} =

A

1p+1\frac{1}{p + 1}

B

11−p\frac{1}{1 - p}

C

1p−1p−1\frac{1}{p} - \frac{1}{p - 1}

D

lim⁡x→0−f(x)=0\lim_{x \rightarrow 0 -}f(x) = 0

Answer

1p+1\frac{1}{p + 1}

Explanation

Solution

lim⁡n→∞1p+2p+3p+.....+npnp+1\lim_{n \rightarrow \infty}\frac{1^{p} + 2^{p} + 3^{p} + ..... + n^{p}}{n^{p + 1}}= lim⁡n→∞∑r=1n[rpnp+1]\lim_{n \rightarrow \infty}\sum_{r = 1}^{n}\left\lbrack \frac{r^{p}}{n^{p + 1}} \right\rbrack

= lim⁡n→∞1n∑r=1n(rn)p=∫01xpdx=[xp+1p+1]01=1p+1\lim_{n \rightarrow \infty}\frac{1}{n}\sum_{r = 1}^{n}\left( \frac{r}{n} \right)^{p} = \int_{0}^{1}{x^{p}dx} = \left\lbrack \frac{x^{p + 1}}{p + 1} \right\rbrack_{0}^{1} = \frac{1}{p + 1}.