Theory behind Integral Test with Riemann Sums

[vanmaiden]

Homework Statement

I've seen two methods that prove the integral test for convergence, but I fear they contradict each other. Each method uses an improper integral where the function f(x) is positive, decreasing, and continuous and f(x) = a_n. What confuses me is one method starts off the riemann sums at n = 1 and makes them all circumscribed rectangles - which make sense to me. The other method starts off the riemann sums at n = 0 and makes them all inscribed rectangles. Can anybody explain to me why each one of these work? Is one right and the other one wrong?

Homework Equations

The Attempt at a Solution

I've reviewed riemann sums in general, but I'm confused as to why two different methods are used in the theory behind the integral test.

 

[vanmaiden]

Just fixed the title

 

[Office_Shredder]

They prove opposite parts of the integral test. When you do a left Riemann sum you get a Riemann sum which overestimates the area - so if the integral diverges (infinite area), the Riemann sum must diverge as well. And if the Riemann sum converges (finite area under the rectangles) the integral must converge as well

On the other hand, if you do a right Riemann sum you get an underestimate for the area under the graph of the function. So if the integral converges (finite area under the graph) you must have a finite Riemann sum as well. And if the Riemann sum diverges, the integral diverges as well.

Since both Riemann sums are the exact same except for the first term of the series, this shows that the series and the integral either both diverge or both converge, you can't have one do one and one do the other

 

[vanmaiden]

They prove opposite parts of the integral test. When you do a left Riemann sum you get a Riemann sum which overestimates the area - so if the integral diverges (infinite area), the Riemann sum must diverge as well. And if the Riemann sum converges (finite area under the rectangles) the integral must converge as well

On the other hand, if you do a right Riemann sum you get an underestimate for the area under the graph of the function. So if the integral converges (finite area under the graph) you must have a finite Riemann sum as well. And if the Riemann sum diverges, the integral diverges as well.

Since both Riemann sums are the exact same except for the first term of the series, this shows that the series and the integral either both diverge or both converge, you can't have one do one and one do the other

I see what you mean, but the two sequences I stated above both start off at n = 1. I understand the idea of shifting and changing the index, but the index stays the same. One source makes the rectangles circumscribed at n = 1 whereas the other makes them inscribed at n = 1. That's pretty much what confuses me.