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.
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.
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?
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.)