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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?

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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.

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  • $\begingroup$ 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. $\endgroup$ – k5f Aug 16 '14 at 17:59
  • $\begingroup$ @Klobbbyyy No, your limit is $-\dfrac{1}{2}$. $\endgroup$ – Hamou Aug 16 '14 at 18:05
  • $\begingroup$ @Klobbbyyy I added some details $\endgroup$ – Petite Etincelle Aug 16 '14 at 18:09
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    $\begingroup$ 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 $-\dfrac{1}{2}$. $\endgroup$ – Hamou Aug 16 '14 at 18:42
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    $\begingroup$ @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. $\endgroup$ – Petite Etincelle Aug 17 '14 at 10:40
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Hint: Use the equivalence $1^p+2^p+\ldots+n^p\sim \dfrac{n^{p+1}}{p+1}$.
$$\dfrac{\sum_{k=1}^nk^p}{n^{p+1}}=\frac{1}{n}\sum_{k=1}^n\left(\dfrac{k}{n}\right)^p=\frac{1-0}{n}\sum_{k=1}^n\left(0+\dfrac{k}{n}(1-0)\right)^p\to\int_0^1x^pdx=\dfrac{1}{p+1}$$

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  • $\begingroup$ What does 'equivalence' mean in this context? $\endgroup$ – k5f Aug 16 '14 at 17:07
  • $\begingroup$ $u_n\sim v_n$ ($v_n\neq 0$)if $\dfrac{u_n}{v_n}\to 1$. $\endgroup$ – Hamou Aug 16 '14 at 17:11
  • $\begingroup$ 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 $\infty-\infty$. $\endgroup$ – k5f Aug 16 '14 at 17:17
  • $\begingroup$ The answer is edited. $\endgroup$ – Hamou Aug 16 '14 at 17:24
  • $\begingroup$ Sorry, I still don't see how it solves the indeterminate form problem. $\endgroup$ – k5f Aug 16 '14 at 17:38

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