1. ## Aternating Series Test

Hello,

As a lemma to proving the "alternating series test" using the Cauchy criterion, I am told I need to show the following, (and then use this fact to prove the AST). (I am told it should be no more than a sentence or two)

Lemma. Let $\displaystyle a_n$ be a positive monotonically decreasing sequence. Suppose that for all $\displaystyle N \in \mathbb{N}$, $\displaystyle m \geq n > N$ implies that $\displaystyle \left| \sum_{k=n}^{m}(-1)^k a_k \right| \leq a_N$. Then $\displaystyle \sum_{k=0}^{\infty} (-1)^ka_k$ satisfies the Cauchy criterion

(A series $\displaystyle \sum a_n$ is said to satsify the cauchy crtierion if and only if for all $\displaystyle \epsilon > 0$, there exists $\displaystyle N$ such that $\displaystyle m \geq n >N$ imply $\displaystyle \left|\sum_{k=n}^{m}a_n\right|< \epsilon$

NOTE: I am not asking for the actual proof of the AST. Just this basic fact above which is a prerequisite to the proof.

Thanks for any help,

James

2. ## Re: Aternating Series Test

Do you know Leibniz convergence criterium?

3. ## Re: Aternating Series Test

No sir I do not. Again, my goal is not to prove the actual AST. I am just trying to justify a preliminary assumption. Perhaps this will make what I want a little clearer.

Statement 1. For all $\displaystyle N \in \mathbb{N}$, $\displaystyle m \geq n > N$ implies $\displaystyle \left|\sum_{k=n}^m(-1)^ka_k\right| \leq a_N$

Statement 2. For all $\displaystyle \epsilon > 0$, there exists $\displaystyle N \in \mathbb{N}$ such that $\displaystyle m \geq n > N$ implies $\displaystyle \left| \sum_{k=n}^m (-1)^k a_k \right| < \epsilon$

I need to show that Statement 1. implies Statement 2. (Prove that Statement 1. is sufficient for Statement 2.). My teacher explained that it should be very short.

Thanks again,

James