Showing a series is less than another

• Dec 2nd 2012, 08:29 PM
zachoon
Showing a series is less than another
I have some issues with this problem.

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

Thanks.
• Dec 2nd 2012, 08:41 PM
Prove It
Re: Showing a series is less than another
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...
• Dec 2nd 2012, 08:44 PM
zachoon
Re: Showing a series is less than another
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.
• Dec 2nd 2012, 08:56 PM
Prove It
Re: Showing a series is less than another
Well in this case, since \displaystyle \begin{align*} a_m < b_m \end{align*}, that means \displaystyle \begin{align*} b_m = a_m + k_m \end{align*} where \displaystyle \begin{align*} k_m > 0 \end{align*}.

That means \displaystyle \begin{align*} \sum b_m = \sum \left( a_m + k_m \right) = \sum a_m + \sum k_m > \sum a_m \end{align*}
• Dec 2nd 2012, 08:58 PM
SworD
Re: Showing a series is less than another
Quote:

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.