Solution July 26, 2007

(Difference between revisions)
 Revision as of 22:33, 26 Jul 2007Landen (Talk | contribs)← Go to previous diff Revision as of 22:34, 26 Jul 2007Landen (Talk | contribs) Go to next diff → Line 6: Line 6: The plan is to use a [http://en.wikipedia.org/wiki/Riemann_integral Riemann sum] to convert the limit into an integral. The plan is to use a [http://en.wikipedia.org/wiki/Riemann_integral Riemann sum] to convert the limit into an integral. - $\lim_{m\rightarrow \infty} \frac{1^n+2^n+\dots+m^n}{m^{n+1}}, = \lim_{m\rightarrow \infty} \frac{1}{m}\sum_{j=1}^m \left(\frac{j}{m}\right)^n=\int_0^1 x^n dx=\frac{1}{n+1}$ + $\lim_{m\rightarrow \infty} \frac{1^n+2^n+\dots+m^n}{m^{n+1}} = \lim_{m\rightarrow \infty} \frac{1}{m}\sum_{j=1}^m \left(\frac{j}{m}\right)^n=\int_0^1 x^n dx=\frac{1}{n+1}$

Revision as of 22:34, 26 Jul 2007

Solution by landen

Find the following limit in terms of $n\,.$

$\lim_{m\rightarrow \infty} \frac{1^n+2^n+\dots+m^n}{m^{n+1}},\ n \in \mathbb{R}$ and $n \in (-1,\infty).\,$

The plan is to use a Riemann sum (http://en.wikipedia.org/wiki/Riemann_integral) to convert the limit into an integral. $\lim_{m\rightarrow \infty} \frac{1^n+2^n+\dots+m^n}{m^{n+1}} = \lim_{m\rightarrow \infty} \frac{1}{m}\sum_{j=1}^m \left(\frac{j}{m}\right)^n=\int_0^1 x^n dx=\frac{1}{n+1}$