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I've calculated $\lim_{n\to\infty}\dfrac{1^p+2^p+\cdots+n^p}{n^{p+1}}=\dfrac1{p+1}$ where $p\in\mathbb{N}$ fixed. I feel it should help me get this one $\lim_{n\to\infty}\left(\dfrac{1^p+2^p+\cdots+n^p}{n^{p}}-\dfrac{n}{p+1}\right)$, but I'm not sure how. Any hints?

Michael Hardy
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k5f
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2 Answers2

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I don't know how to get the second limit by using the first one since $$\lim_{n\to\infty}\left(\dfrac{1^p+2^p+\ldots+n^p}{n^{p}}-\dfrac{n}{p+1}\right) = \lim_{n\to\infty}n\left(\dfrac{1^p+2^p+\ldots+n^p}{n^{p+1}}-\dfrac{1}{p+1}\right)$$

which is of $0\times\infty$ type.

But both limits can be computed using Stolz–Cesàro theorem through binomial formula

Added: \begin{align} &\lim_{n\to\infty}\left(\dfrac{1^p+2^p+\ldots+n^p}{n^{p}}-\dfrac{n}{p+1}\right)\\ =&\lim_{n\to\infty}\left(\dfrac{(p+1)(1^p+2^p+\ldots+n^p)-n^{p+1}}{(p+1)n^{p}}\right) \\ =&\lim_{n\to\infty}\left(\dfrac{(p+1)(n+1)^p-(n+1)^{p+1}+n^{p+1}}{(p+1)((n+1)^{p}-n^p)}\right)\\ =&\lim_{n\to\infty}\left(\dfrac{\frac{(p+1)p}{2}n^{p-1}+\text{term of lower order}}{(p+1)pn^{p-1} + \text{term of lower order}}\right)\\ =& \frac{1}{2} \end{align}

In the comment, @Hamou gives a way to get directly the limit by application of theorem on convergence rate of Riemann sum. For exact statement of the theorem, look into the reference therein.

Petite Etincelle
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  • Thanks! I used the theorem and then $p$ times L'Hospital and got the answer $\frac{1-n}{n}$. This however appears to be wrong, as the result is supposed to be $-\infty$, says Mathematica software. – k5f Aug 16 '14 at 17:59
  • @Klobbbyyy No, your limit is $-\dfrac{1}{2}$. – Hamou Aug 16 '14 at 18:05
  • @Klobbbyyy I added some details – Petite Etincelle Aug 16 '14 at 18:09
  • @Hamou Thank you! Just realized my mistake. – k5f Aug 16 '14 at 18:15
  • @LiuGang Could you please explain where did the fourth line came from? – k5f Aug 16 '14 at 18:22
  • @Klobbbyyy It's by binomial formula, see here. – Petite Etincelle Aug 16 '14 at 18:24
  • @LiuGang Solved, many thanks. – k5f Aug 16 '14 at 18:35
  • @LiuGang Everything is right until the last 2 equalities... You can see that the answer is $-\inf$ – shooting-squirrel Aug 16 '14 at 18:35
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    I've two remarks: the limit is negative and why the $b_n$'s sequence is strictly monotone. If you try by using the approximate theorem of Riemann sum: On the subspace $\mathcal{C}^1([a,b])$ of continuously differentiable functions, we have

    $$\lim_{n\to\infty} \sum_{k=1}^{n} f\left(a+\frac{k}{n}(b-a)\right) - n\int_a^b f(x),dx = \frac{f(a) - f(b)}{2}.$$ here $a=0$ and $b=1$ and $f(x)=x^p$. we get the limit as $-\dfrac{1}{2}$.

     – Hamou Aug 16 '14 at 18:42
  • @Hamou +1 for the theorem, but take $f=x, a=0, b = 1$ in your theorem to see the sign – Petite Etincelle Aug 16 '14 at 18:59
  • @LiuGang Thank you for your remark, the formula is: $$\lim_{n\to\infty} \sum_{k=0}^{n-1} f\left(a+\frac{k}{n}(b-a)\right) - n\int_a^b f(x),dx = \frac{f(a) - f(b)}{2}.$$ – Hamou Aug 16 '14 at 19:08
  • @Hamou Could you give a reference for this theorem? – Petite Etincelle Aug 16 '14 at 19:30
  • See page $478$ in this Document. – Hamou Aug 16 '14 at 19:34
  • @Hamou Thank you very much – Petite Etincelle Aug 16 '14 at 19:53
  • @Hamou Could you please explain what's wrong with the calculation? Why doesn't the Stolz theorem work? $(p+1)n^p$ is strictly increasing and goes to $\infty$, as the theorem asks it to be. – k5f Aug 17 '14 at 10:27
  • @Klobbbyyy There's no problem with the calculation, see the Document given by Hamou, that's a different application. – Petite Etincelle Aug 17 '14 at 10:33
  • @LiuGang But he claims the limit is $-\frac1{2}$, not $\frac1{2}$. I'm still a beginner, so please bear with me. – k5f Aug 17 '14 at 10:36
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    @Klobbbyyy that's for the limit of $\left(\dfrac{1^p+2^p+\ldots+(n-1)^p}{n^{p}}-\dfrac{n}{p+1}\right)$(which is your limit minus 1). Btw, Hamou's citation is not exact, look into his document for exact statement of the theorem. – Petite Etincelle Aug 17 '14 at 10:40
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Hint: Use the equivalence $1^p+2^p+\ldots+n^p\sim \dfrac{n^{p+1}}{p+1}$.
$$\dfrac{\sum_{k=1}^nk^p}{n^{p+1}}=\frac{1}{n}\sum_{k=1}^n\left(\dfrac{k}{n}\right)^p=\frac{1-0}{n}\sum_{k=1}^n\left(0+\dfrac{k}{n}(1-0)\right)^p\to\int_0^1x^pdx=\dfrac{1}{p+1}$$

Hamou
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  • What does 'equivalence' mean in this context? – k5f Aug 16 '14 at 17:07
  • $u_n\sim v_n$ ($v_n\neq 0$)if $\dfrac{u_n}{v_n}\to 1$. – Hamou Aug 16 '14 at 17:11
  • I have a problem with this. If I am to plug this in, I first have to break the limit down using limit laws. But to get to $1^p+2^p+\ldots+n^p$ in order to use the equivalence, I would have to pass through indeterminate forms like $\infty-\infty$. – k5f Aug 16 '14 at 17:17
  • The answer is edited. – Hamou Aug 16 '14 at 17:24
  • Sorry, I still don't see how it solves the indeterminate form problem. – k5f Aug 16 '14 at 17:38