Question: > 
> The alternating series $\Sigma(-1)^{k+1}a\_k$ converges provided:
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> 
> 1) $0< a\_{k+1} \le a\_k$ for $k$ greater than some index $N$.
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> 2) $\lim\_{k\to \infty}a\_k = 0$
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> 
> 

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?

Answer: 
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.