# Math Help - Riemann sum.

1. ## Riemann sum.

$\displaystyle \lim_{x\rightarrow \infty}\sum_{k=1}^{x}\frac{(k-1)^4}{x^5}$

I'm sure this is a integral of some sort but I can not get it into the correct format to figure out the limits of integration (or the integrand! ):

$\displaystyle \int_{a}^{b}f(t)dt=\lim_{x\rightarrow \infty}\sum_{k=1}^{x} \left ( \frac{b-a}{x}\right )\cdot f\left ( a+k\frac{b-a}{x} \right )$

edit: sorry, typo, I changed it.

2. We may write

$
\displaystyle
\sum \frac{1}{x} \; \frac{(k-1)^4}{x^4}=\sum \frac{1}{x} \; (\frac{k-1}{x})^4
$

3. That is where I got originally. I simply do not know what to do from there.

4. We have $\displaystyle \sum_{k=1}^x\frac 1x\left(\frac{k-1}x\right)^4=\sum_{k=2}^x\frac 1x\left(\frac{k-1}x\right)^4=\sum_{j=1}^{x-1}\frac 1x\left(\frac jx\right)^4 =\sum_{j=1}^x\frac 1x\left(\frac jx\right)^4-\frac 1x$ and now you can conclude.

5. So:

$\displaystyle \sum_{i=1}^{n}\frac{(i-1)^4}{n^5}$

$=\displaystyle \sum_{i=1}^{n}\left (\frac{i-1}{n} \right )^4 \frac{1}{n}$

$=\displaystyle \sum_{J=2}^{n-1}\left ( J\frac{1}{n} \right )^4 \frac{1}{n}$

$\therefore a=0, b=1$

$\displaystyle \sum_{i=1}^{n}\frac{(i-1)^4}{n^5}= \sum_{J=2}^{n-1}\left ( J\frac{1}{n} \right )^4 \frac{1}{n}=\int_{0}^{1}x^4dx=\frac{1}{5}$

?

6. Originally Posted by integral
So

$=\displaystyle \sum_{J=2}^{n-1}\left ( J\frac{1}{n} \right )^4 \frac{1}{n}$
I think you have to start the sum from $J=1$ (but it doesn't change anything because the result you found is correct).