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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?


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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 Feb 7 '12 at 2:35
Do you mean $$\lim_{n\to\infty}\frac1{n^{p+1}}\sum_{i=1}^ni^p\;?$$ – Brian M. Scott Feb 7 '12 at 2:36


$$\frac{1}{n^{p+1}} k^p= \frac{1}{n} \left(\frac{k}{n}\right)^p. $$

Now what would a Riemann sum of $\int_0^1 x^p dx$ look like, with $[0,1]$ divided into $n$ equal pieces?

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Thanks! That was so trivial! :) – Jr. Feb 7 '12 at 4:06

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.)

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