Exponential limits

Calculus Level 2

lim n 1 p + 2 p + 3 p + + n p n p + 1 = ? \large \lim_{n \to \infty} \frac {1^p+2^p+3^p + \cdots + n^p}{n^{p+1}} = ?

1 p + 2 \dfrac{1}{p+2} 1 p 1 \dfrac{1}{p-1} 1 p 1 p 1 \dfrac{1}{p} - \dfrac{1}{p-1} 1 p + 1 \dfrac{1}{p+1}

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1 solution

Chew-Seong Cheong
Nov 17, 2017

Relevant wiki: Riemann Sums

L = lim n 1 p + 2 p + 3 p + + n p n p + 1 = lim n k = 1 n k p n p + 1 = lim n 1 n k = 1 n ( k n ) p By Riemann sums = 0 1 x p d x = x p + 1 p + 1 0 1 = 1 p + 1 \begin{aligned} L & = \lim_{n \to \infty} \frac {1^p+2^p+3^p + \cdots + n^p}{n^{p+1}} \\ & = \lim_{n \to \infty} \sum_{k=1}^n \frac {k^p}{n^{p+1}} \\ & = \lim_{n \to \infty} \frac 1n \sum_{k=1}^n \left(\frac kn\right)^p & \small \color{#3D99F6} \text{By Riemann sums} \\ & = \int_0^1 x^p dx \\ & = \frac {x^{p+1}}{p+1} \ \bigg|_0^1 \\ & = \boxed{\dfrac 1{p+1}} \end{aligned}

Thank you Sir

Tushar Maske - 3 years, 6 months ago

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