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Find the limit of the following sequence:

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

I've tried using the Stolz-Cesaro theorem. I've set $a_n=1^p+2^p+\cdots+n^p$, and $b_n=n^{p+1}$.

After some transformations I got $$\lim_{{n}\to{\infty}} \frac{(n+1)^p}{(n+1)^p-n^p},$$ but after expanding the binomials, I could not make any transformations that would give me a satisfactory result. Can you help me please? Thank you very much.

Daniel W. Farlow
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wonderingdev
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    You should get $\frac{(n+1)^p}{(n+1)^{\mathbf{p+1}}-n^{\mathbf{p}}}$ $=\frac{(n+1)^p}{(p+1)n^p+\text{lower order terms}}=\frac{(1+1/n)^p}{(p+1)+\text{ terms that tend to zero}}\to\frac{1}{p+1}$. – Pp.. Jan 18 '15 at 07:47
  • Related: http://math.stackexchange.com/questions/948506/evaluating-lim-n-to-infty-frac199-299-cdots-n99n100 – taninamdar Jan 18 '15 at 07:56
  • Thank you for your suggestions – wonderingdev Jan 18 '15 at 08:26

2 Answers2

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Hint: $S(n,p) = \dfrac{1}{n}\left(\left(\dfrac{1}{n}\right)^p + \left(\dfrac{2}{n}\right)^p+\cdots \left(\dfrac{n}{n}\right)^p\right)\to \displaystyle \int_{0}^1 x^pdx$

DeepSea
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  • But is there any other possibilities, because I've never worked with integrals when calculating limits? In my book is writen to use Stolz-Cesaro. – wonderingdev Jan 18 '15 at 07:39
  • @JohnG. Then use it. It is also direct that way. In the fraction you have, simplify the denominator (notice the exponents should be $p+1$ not $p$), then divide numerator and denominator by $n^p$. – Pp.. Jan 18 '15 at 07:42
  • Yes, I showed what I've done, but I could not simplify the expression when expanding binomials. – wonderingdev Jan 18 '15 at 07:44
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Applying Stolz-Cesaro leads to:

$$\lim_{{n}\to{\infty}} \frac{(n+1)^p}{(n+1)^{p+1}-n^{p+1}} = \lim_{{n}\to{\infty}} \frac{1}{n\left(1+\frac{1}{n}\right)^{p+1}-n}$$

dividing numerator and denominator by $n^{p}$.

Expand the denominator in power series:

$$n\left(1+\frac{1}{n}\right)^{p+1}-n = \binom{p+1}{1} + \binom{p+1}{2}\frac{1}{n}+\mathcal{O}(1/n^2)$$

Clearly it approaches $p+1$ as $n \to \infty$.

Hence, your required limit is $\frac{1}{p+1}$

If you are looking for slightly different arguments apart from the Stolz-Cesaro or the already mentioned idea in the other answer,

Here's a nice one by @Jack D'Aurizio

$\displaystyle k(k-1)\cdot\ldots\cdot(k-p+1)=p!\binom{k}{p}\leq k^p \leq p!\binom{k+p}{p}=(k+p)\cdot\ldots\cdot(k+1)$

for each $1 \le k \le n$

Thus,

$$p!\binom{n+1}{p+1} = p!\sum\limits_{k=1}^{n}\binom{k}{p}\leq \sum\limits_{k=1}^{n} k^p \leq p!\sum\limits_{k=1}^{n}\binom{k+p}{p} = p!\binom{n+p+1}{p+1}$$

Thus, $\displaystyle \lim_{{n}\to {\infty}}\frac{1^p+2^p+\cdots+n^p}{n^{p+1}}= \frac{1}{p+1}$ by Squeeze Theorem.

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