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$$\lim_{n\to\infty} \dfrac{1^p+2^p+...+n^p}{n^p}-\frac{n}{p+1}\text{ with } p\in\mathbb N$$

I don't really know how to start, I mean... I could try substituting by $\Big(\dfrac{n(n+1)}{2}\Big)^p$ but that would do me anything? Any hints?

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marked as duplicate by Paramanand Singh limits Nov 11 '18 at 15:13

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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HINT

We can use Cesàro-Stolz to

$$(p+1)\frac{1^p+2^p+...+n^p-\frac{n^{p+1}}{p+1}}{(p+1)n^p}$$

to obtain

$$\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=(p+1)\frac{(n+1)^p-\frac{(n+1)^{p+1}}{p+1}+\frac{n^{p+1}}{p+1}}{(p+1)(n+1)^p-(p+1)n^p}=$$

$$\frac{(p+1)(n+1)^p-(n+1)^{p+1}+n^{p+1}}{(p+1)(n+1)^p-(p+1)n^p}$$

and then use the binomial theorem.

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  • $\begingroup$ how to use the binomial theorem then? $\endgroup$ – iggykimi Nov 11 '18 at 15:45
  • $\begingroup$ For example in the denominator we can use that $(n+1)^p=n^p+pn^{p-1}+...$ and we have some cancellation, therefore the $p(p+1)n{^p-1}$ term dominates, and so on for others term. $\endgroup$ – gimusi Nov 11 '18 at 16:07
  • $\begingroup$ @iggykimi errata corrige: the dominant term at the denominator is of course $p(p-1)n^{p-1}$. $\endgroup$ – gimusi Nov 11 '18 at 16:15
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If you know generalized harmonic numbers $$\sum_{k=1}^n k^p=H_n^{(-p)}$$ and their asymptotics $$H_n^{(-p)}=n^p \left(\frac{n}{p+1}+\frac{1}{2}+\frac{p}{12 n}+O\left(\frac{1}{n^3}\right)\right)+\zeta (-p)$$ making $$\frac{\sum_{k=1}^n k^p }{n^p }-\frac n {p+1}=\frac{1}{2}+\frac{p}{12 n}+\cdots$$ which shows the limit and how it is approached.

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