Thread: Showing a series is less than another

I have some issues with this problem.

Let (the sum of a_n from n=1 to infinity) and (the sum of b_n from n=1 to infinity) be two convergent series. If a_n (less than or equal to) b_n for all n in Natural Number, and a_m < b_m for some m in Natural Number, show that (the sum of a_n from n=1 to infinity) < (the sum of b_n from n=1 to infinity).

Thanks.

Sorry but isn't this statement obvious? If all terms in the a series are less than the terms in the b series, then surely the a series must be smaller than the b series...

Well, ish, but a lot of the elementary analysis I'm dealing with seems to be proving somewhat obvious things with a more roundabout method. I mean, if a problem like this came up on the test, I couldn't really just state the statement is self-explanatory. But for me, I have more of an issue of figuring out how to start my proofs.

Well in this case, since , that means where .

That means

Originally Posted by Prove It

Sorry but isn't this statement obvious? If all terms in the a series are less than the terms in the b series, then surely the a series must be smaller than the b series...

Depends on who it's obvious to. Half of the things that are "proven" these days as exercises for relevant classes are obvious. The demand for rigor never seems to end.

Well I guess you could argue that a_n will in at least once instance be less than b_n, and then will never possibly exceed or equal b_n, therefore the sum is smaller... depends what kind of arguments you are allowed to use.