How do I show that :
$$\lim_{n\to\infty} \frac{1}{n^{p+1}} \sum_{i=1}^{n} i^p = \frac{1}{p+1}$$
using a Riemann sum?
Thanks.
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Did you copy the problem right? Your limit takes x going to infinity, but x doesn't appear inside the limit at all. Do you mean the limit as n-> infinity? – Ben FL Feb 07 '12 at 02:35
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Do you mean $$\lim_{n\to\infty}\frac1{n^{p+1}}\sum_{i=1}^ni^p;?$$ – Brian M. Scott Feb 07 '12 at 02:36
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Does this answer your question? Prove that $\lim_{n\rightarrow\infty}\frac{1}{n^{p+1}}\sum_{k=1}^{n}k^{p}=\frac{1}{p+1} $ – Sil Nov 22 '22 at 04:43
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Observe
$$\frac{1}{n^{p+1}} k^p= \frac{1}{n} \left(\frac{k}{n}\right)^p. $$
Now what would a Riemann sum of $\int_0^1 x^p dx$ look like, with $[0,1]$ divided into $n$ equal pieces?
anon
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Assuming that you meant $$\lim_{n\to\infty}\frac1{n^{p+1}}\sum_{i=1}^ni^p=\frac1{p+1}\;,$$ compare $\displaystyle\frac1{n^{p+1}}\sum_{i=1}^ni^p$ with the integrals $$\frac1{n^{p+1}}\int_0^n x^pdx\quad\text{and}\quad\frac1{n^{p+1}}\int_1^{n+1}x^pdx\;;$$ a sketch should help greatly. (If you use this approach, you’ll need to split it into two cases, $p\ge 0$ and $p<0$, but the two are handled almost identically.)
Brian M. Scott
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