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The limit the following sequence is $\frac{1}{p+1}$. I have tried to prove it, but I don't success. Could you help me?

$$x_n=\frac{1^p+2^p+...+n^p}{n^{p+1}}$$

$$p\in\mathbb{N}$$

Thanks in advance!

Gibbs
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4 Answers4

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hint

$$x_n=\frac 1n\sum_{k=1}^nf(\frac kn)$$

with $$f:t\mapsto t^p$$

$x_n$ is a Riemann sum of $f$ which is integrable. thus $$\lim_{n\to+\infty}x_n=\int_0^1t^pdt=\frac{1}{p+1}$$

hamam_Abdallah
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HINT

By Stolz-Cesaro

$$\lim_{n \to \infty} \frac{a_n}{b_n} =\lim_{n \to \infty} \frac{a_{n+1}-a_n}{b_{n+1}-b_n} $$

that is

$$\lim_{n \to \infty} \frac{1^p+2^p+...+n^p}{n^{p+1}} =\lim_{n \to \infty} \frac{(n+1)^{p}}{(n+1)^{p+1}-n^{p+1}} $$

user
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  • So now I think I should develop $(n+1)^{p+1}$ in order to cancel $n^{p+1}$ – Gibbs Oct 17 '18 at 21:19
  • @P4U That's a nice guess :) – user Oct 17 '18 at 21:19
  • $$\lim_{n \to \infty}\frac{(n+1)^p}{n^{p+1}+(p+1)n^p+(\frac{p^2+1}{2})n^{p-1}+...-n^{p+1}}$$

    Then is it correct to divide numerator and denominator by $n^p$?

     – Gibbs Oct 17 '18 at 21:28
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    @P4U More simply note that by binomial theorem for the numerator we have $(n+1)^p \sim n^p$ and the denominator $(n+1)^{p+1}-n^{p+1}\sim n^{p+1}+(p+1)n^p-n^{p+1}=(p+1)n^p$. – user Oct 17 '18 at 21:30
  • True, I didn't notice that. Thank you!! – Gibbs Oct 17 '18 at 21:34
  • @P4U That's nice! You were almost there. Bye – user Oct 17 '18 at 21:35
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Using the definition of generalized harmonic numbers $$\sum_{i=1}^n i^p=H_n^{(-p)}$$ Now, using the asymptotics $$H_n^{(-p)}=n^p \left(\frac{n}{p+1}+\frac{1}{2}+\frac{p}{12 n}+O\left(\frac{1}{n^3}\right)\right)+\zeta (-p)$$ $$\frac{\sum_{i=1}^n i^p } {n^{p+1}}=n^{-(p+1)} \zeta (-p)+\left(\frac{1}{p+1}+\frac{1}{2 n}+\frac{p}{12 n^2}+O\left(\frac{1}{n^4}\right)\right)=\frac{1}{p+1}+\frac{1}{2 n}+O\left(\frac{1}{n^2}\right)$$

Use it for $n=10$ and $p=9$. The exact result would be $\frac{314860997}{2000000000}\approx 0.15743$ while the last approximation would give $\frac 3{20}=0.15$.

Claude Leibovici
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Since $p\in\mathbb {N} $ it is instructive to prove the result via induction. The result clearly holds for $p=0$. Let's assume that it holds for $p=0,1,2,\dots,m-1$ and then consider the case for $p=m$. Let $S_m(n)=\sum_{i=1}^{n}i^m$. We have $$n^{m+1}-(n-1)^{m+1}=\binom{m+1}{1}n^m-\binom{m+1}{2}n^{m-1}+\dots$$ And summing over index $n$ from $1$ to $n$ we get $$n^{m+1}=(m+1)S_m(n) - \binom{m+1}{2}S_{m-1}(n)+\dots $$ Dividing by $n^{m+1}$ we get $$1=(m+1)\frac{S_m(n)}{n^{m+1}}-\binom {m+1}{2}\frac{S_{m-1}(n)}{n^{m+1}}+\dots$$ Taking limits as $n\to\infty$ we get via our induction hypotheses $$1=(m+1)\lim_{n\to\infty}\frac{S_{m}(n)}{n^{m+1}}$$ as rest of the terms in RHS tend to $0$. Thus we have $$\lim_{n\to\infty} \frac{1^m+2^m+\dots+n^m}{n^{m+1}}=\frac{1}{m+1}$$ Thus the result holds for $p=m$ also and the proof is now complete by induction.

Paramanand Singh
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