# Compare a series sum to it's equivelent integral

1. Sep 24, 2007

### stufletcher

1. The problem statement, all variables and given/known data

excuse my formatting.

compare

sum(from n=2 -> infinity) of 5/(n^1.1)

to

integral (from 1 -> infinity) of 5/(x^1.1)

2. Relevant equations

3. The attempt at a solution

if it was sum and integrate from 2 it would be easy... the sum is a rienmann sum using right endpoints and delta-x of 1, and it's error will always be larger than it's integral on a decreasing curve.

however because the function is decreasing, and the integral starts at 1, not 2, i think i have to calculate the size of the error of the rienmann compared to it's integral from 2 -> infinity, and compare that to the integral from 1 ->2.

maybe i'm just having a brain coniption, but i cant seem to sum either the rienmann or the integral from 2 -> infinity... anyone have any suggestions?

thanks

stu

2. Sep 24, 2007

### genneth

It's easier to see what to do with a simpler function. Let's use 1/x.

Draw a graph, using bars of width 1 to represent each value, so you end up with a bar graph of 1/x. The sum of the area under the bars is the sum of 1/x. Now notice that you can have two curves, one joining up the top left corners of the all bars, and one joining up the top right corners. These curves are 1/x curves, but one of them is shifted along by 1 unit. The area under these curves can be found by integrating 1/x, between appropriate limits. Can you see how to apply this method to your problem?

3. Sep 24, 2007

### stufletcher

it's always something simple isnt it..

area $$\frac{5}{x^{1.1}} * (x - (x-1)) < \int_{x-1}^x\frac{5}{x^{1.1}}dx$$

for all values of positive x, so the sum must be smaller than the integral

thanks for the tip

stu

Last edited: Sep 24, 2007