# Alternating Series?

The alternating series $\Sigma(-1)^{k+1}a_k$ converges provided:

1) $0< a_{k+1} \le a_k$ for $k$ greater than some index $N$.

2) $\lim_{k\to \infty}a_k = 0$

I'm confused about the language. Is the theorem implying that if the conditions are not met, then the series is necessarily divergent? Or is it stating that if the conditions are met then the series if necessarily convergent, but if there not then the series might be convergent or divergent?

• what do you mean by the statements such as "if A then B" ? – wanderer Mar 18 '14 at 19:43
• @wanderer $A \implies B$ I think. I haven't worked with logic before. – dfg Mar 18 '14 at 19:44

You are asking about the converse, and the converse to this statement is false. The converse is that if $\sum (-1)^{k+1}a_k$ is alternating, and either $\displaystyle\lim_{n \to \infty}a_k \neq 0$ or $a_k$ is not monotonic decreasing and positive, then the series diverges. It is certainly true that if $\displaystyle\lim_{n \to \infty}a_k \neq 0$ then the series diverges. However, there are convergent alternating series for which $a_k$ is not monotonic decreasing and $\displaystyle\lim_{n \to \infty}a_k = 0$.

For example, consider the series $\displaystyle\sum_{k=1}^\infty b_k = \frac{1}{2^2} - \frac{1}{1^2} + \frac{1}{4^2} - \frac{1}{3^2} + \frac{1}{6^2} - \frac{1}{5^2} + \cdots$. This series is a rearrangement of the series $\displaystyle\sum_{k=1}^\infty \frac{(-1)^k}{k^2}$, which is absolutely convergent.

Statements such as $P\Rightarrow Q$ in words mean that if $P$ is true then so is $Q$. It doesn't mean that $Q$ is true "only if" $P$ is true. $P\Rightarrow Q$ is same as $\sim Q\Rightarrow \sim P$, i.e., if $Q$ is not true then $P$ can't be true.

Hope this helps.

• Thanks, but I'm not entirely sure if $A \implies B$ is what the theorem is saying. I guess what I'm asking is if the theorem is saying: conditions are not met $\implies$ divergence. – dfg Mar 18 '14 at 19:53
• Not neccesarily...e.g. you can interchange the following terms of the series $1/n^{2}$ and $1/n^{3}$...then the first condition will be failed...but the series again converges to the same limit. – wanderer Mar 18 '14 at 19:57

The statement neither uses the logical implication sign nor "if... then..." language, so I think the confusion has not been addressed. What it does say is "X provided (A and B)". This means the statement X is true when you have determined both A and B as true. X is the statement "the alternating series... converges" and A and B are the clauses 1 and 2. We know that there is an implied "and" here between A and B because of the conventions of English and the use of "provided" without explicitly saying "or" between them.

To flip this around into the logical terms being discussed, this can be written as:

$$(A \wedge B) \implies X$$

Meaning that - as stated - this tells us we know X is true when A and B are both true, but we don't know whether X is true or false otherwise. However, can you say more? Just understanding that the logical view of the statement does not make the claim doesn't mean the claim is false. In fact, can you think of any convergent series where either A or B fail? What is actually proven in the text?

The point is that sometimes a theorem is stated loosely with language and you have to be careful to understand the English connotations. It is actually true that the series will diverge in other conditions. Sometimes, in English, we use the term "provided" to "loosely mean" if and only if. By "loosely mean" I just mean that this is not the textbook meaning, it is not how it is always or regularly employed, but sometimes in certain formulations, it is kind of understood. I think this is probably one of those uses, it has that kind of feel, if you were to bring it up to a teacher, they might explain that meaning, but it is kind of sloppy and unfair to infer that when it is not the real meaning of what is said. It should be corrected if the text actually proves the stronger result.