limn→∞ 1p+2p+3p+…+npnp+1,p>−1 is

limn→∞ 1n∑r=1n rnp=∫01 xpdx=xp+1p+10l=1p+1

limn→∞ 1p+2p+3p+…+npnp+1,p>−1 is

limn→∞ 1n∑r=1n rnp=∫01 xpdx=xp+1p+10l=1p+1

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$lim_{n \to \infty} \frac{1^{p} + 2^{p} + 3^{p} + \dots + n^{p}}{n^{p + 1}}, p > - 1$ is

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

$\frac{1}{p - 1}$

$\frac{1}{p + 2}$

$\frac{1}{p (1 - p)}$

answer is A.

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$lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{n} {(\frac{r}{n})}^{p} = \int_{0}^{1} x^{p} dx = {\frac{x^{p + 1}}{p + 1}|}_{0}^{l} = \frac{1}{p + 1}$

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limn→∞ 1p+2p+3p+…+npnp+1,p>−1 is

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$lim_{n \to \infty} \frac{1^{p} + 2^{p} + 3^{p} + \dots + n^{p}}{n^{p + 1}}, p > - 1$ is