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Have you any idea how to find the limit of the following sum: $$\lim_{n \to \infty}\frac{1^p+2^p+\ldots+n^{p}}{n^{p+1}}.$$

Stolz-Cesaro? any more ideas?

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2 Answers 2

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The quickest way is using integral: $$ \lim \sum_{k=1}^n \frac{k^p}{n^p}\frac{1}{n}=\int_0^1 x^pdx=\frac{1}{p+1}. $$ Seeing this, you may find an elementary proof from $$ k^p\sim \frac{1}{p+1}\bigg[(k+1)^{p+1}-k^{p+1}\bigg]. $$

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    $\begingroup$ That's very nice! $\endgroup$
    – JavaMan
    Commented Jun 13, 2013 at 21:35
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Prove by induction on $p$ that $$f_p(n)=\sum_{k=0}^n k^p$$

is a polynomial of degree $p+1$ with leading term $$\frac{n^{p+1}}{p+1}$$

Immediately conclude the limit is $$\frac{1}{p+1}$$


Alternatively, use Stolz Cesaro, to show your limit is the same as $$\lim\limits_{n\to\infty}\frac{(n+1)^{p}}{(n+1)^{p+1}-n^{p+1}}$$

Then use that $$\lim_{x\to 0}\frac{(1+x)^{\alpha}-1}x=\alpha$$

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