# Proof of Inequality Between Lower and Upper Bounds

1. Jul 10, 2014

### muzak

Convergence of Divergent Series Whose Sequence Has a Limit

1. The problem statement, all variables and given/known data
Suppose $∑a_{n}$ is a series with lim $a_{n}$ = L ≠ 0. Obviously this diverges since L ≠ 0. Suppose we make the new series, $∑(a_{n} - L)$. My question is this: is there some sufficient condition we could put solely on $a_{n}$ (or maybe its sum) without calculating the limit or referencing it other than the existence of a limit so that the new series converges?

2. Relevant equations
I can't think of anything here; I could list all of the convergence tests from Ch. 3 of Baby Rudin but doesn't seem wholly relevant with the way I phrased the question. Feels like this is probably something trivial from some higher level analysis course or analytic number theory course that I haven't had the chance to take yet.

There's one thing that seems possibly relevant (from Rudin):
Theorem Suppose $a_{1}≥a_{2}≥a_{3}≥...≥0.$ Then the series $∑a_{n}$ converges if and only if the series $∑2^{k}a_{2^{k}}$ converges.

3. The attempt at a solution
I really don't know how to approach this without introducing the L somehow into the proof which isn't what I really wish to explore.

I thought to make some simple conditions first to try and reduce the nature of the problem, like making the sequence monotonic and non-negative. With the theorem, I was thinking of maybe introducing some sort of scaling maybe with the $2^{k}$ factor? I don't know.

I think the root test gives convergence for 0 < L < 1, but what about for L > 1? But this sort of references the limit.

Last edited: Jul 10, 2014
2. Jul 10, 2014

### micromass

Staff Emeritus
That's a pretty vague question. Is this an actual book problem or a question you invented yourself? I don't see any reason why there would be such a simple condition. Well, obviously you can try all the series tests you've seen and adapt it to this case. But I don't really know what kind of answer you want here.

3. Jul 10, 2014

### muzak

Invented. Just curious if there exists any such condition to give convergence. I don't know anything about techniques to deal with divergent series, so I'd be satisfied with any specific reference material in lieu of some answer.

4. Jul 10, 2014

### micromass

Staff Emeritus
5. Jul 10, 2014

### muzak

Alright, thanks. Messed up the title of this thread but think I corrected it. Guess I can't fix the thread title, ah well.

Criterion of Abel looks promising maybe:
The criterion of Abel
Let ∑+∞n=0an be a (real or complex) convergent series . Let (vn)n be a bounded sequence of real numbers which is either nondecreasing or nonincreasing. Then the sequence ∑+∞n=0vnan converges.

$v_{n}$ could be the $a_{n}$ and would have to figure out some convergent sequence so that the product equals (L - $a_{n}$).

Last edited: Jul 10, 2014