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I have$$\lim_{n\to \infty} \frac{1^p+2^p+\ldots+n^p}{n^{p+1}}=$$ I managed to simplify it down to $$=\lim_{n\to \infty}\left( \left(\frac1n\right)^p \cdot \frac1n + \left(\frac2n\right)^p \cdot \frac1n + \ldots + \left(\frac{n}{n} \right)^p \cdot \frac1n \right)=$$ $$=\lim_{n\to \infty} \sum_{i=1}^{n} \left(\frac{i}{n}\right)^p \frac{1}n$$ How should I calculate this limit?

Martin Sleziak
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vilbur
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  • Can you express that limit as an integral (as the title of your question seems to imply)? – Prahlad Vaidyanathan Dec 02 '13 at 07:09
  • See http://math.stackexchange.com/questions/478344/what-is-the-result-of-lim-n-to-infty-frac-sumn-i-1-iknk1 and http://math.stackexchange.com/questions/150391/evaluate-lim-limits-n-to-infty-frac-sum-k-1n-kmnm1 – Martin Sleziak Sep 19 '14 at 11:46

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Let us denote $1^p + 2^p + \cdots n^p = P_p(n)$. This is a polynomial of degree $p+1$ and is given by $$P_p(n) = \frac1{p+1} \sum_{k=0}^p \dbinom{p+1}{k} B_k n^{p+1-k}$$ where $B_k$ are the Bernoulli numbers. These polynomials are related to the Bernoulli polynomials and there are some really nice results on these polynomials and more can be found here.

Hence, $$\dfrac{P_p(n)}{n^{p+1}} = \dfrac1{p+1} \sum_{k=0}^p \dbinom{p+1}{k} B_k n^{-k} = \dfrac1{p+1} \left(B_0 + \mathcal{O} \left(\frac1{n}\right) \right)$$ where $B_0 = 1$, $B_1 = \frac12$, etc. Now conclude what you want.

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Community
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Use $$\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$

to $$=\lim_{n\to \infty} \frac1n\sum_{i=1}^n \left(\frac in\right)^p$$ here $\displaystyle f\left(\frac in\right)=\left(\frac in\right)^p$

lab bhattacharjee
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