Trouble understanding divergence of alternating sequences I started studying sequences and series only a few days ago using Stewart's Early transcendentals. I am having trouble understanding how to prove the divergence of alternating series via proving the divergence of the sequence.
Suppose we have $\sum (-1)^{n}b_n$ and $\lim_{n\to\infty} b_n\neq0$. This means the series does not pass the alternating series test, but it should not mean the series necessarily diverges. However, I found online (e.g., here) that it suffices to show that $\lim_{n\to\infty} |a_n|\neq0$ to prove the divergence of the sequence (and therefore the divergence of the series).
For a general alteranting series $a_n$, I do not understand why $\lim_{n\to\infty} |a_n|\neq0$ proves the alternating sequence diverges. In Stewart's book, it is only stated that $\lim_{n\to\infty} |a_n|=0 \implies \lim_{n\to\infty} a_n = 0$, which of course does not translate into saying that $ \lim_{n\to\infty} |a_n| \neq 0$ implies the alternating sequence diverges.
An example of a problem where I struggled with this confussion.
For example, I was asked to prove whether $\sum_{n=1}^\infty \frac{(-2)^n}{n^2}$ converges absolutely, conditionally or diverges. It is clear, via double application of L'Hopital's rule, that $\lim_{x\to\infty} \frac{2^x}{x^2}$ diverges and therefore the series is not absolutely convergent.
Now, to show it diverges conditionally, what I would do given my current understanding is to show, where $a_n= \frac{(-2)^n}{n^2}$, that $a_{2n}$ and $a_{2n-1}$ converge to different limits (or don't exist), implying the sequence (and therefore the series) diverges. However, according to the referenced online sources, it suffices to show that $\lim_{n\to\infty} |a_n| \neq 0$, which was already done to prove the series is not absolutely convergent.
Again, I don't see why  $\lim_{n\to\infty} |a_n| \neq 0$ would implie the sequence (and therefore the series) diverges, and the only similar theorem I know is the one that states that $\lim_{n\to\infty} |a_n|=0 \implies \lim_{n\to\infty} a_n = 0$.
Note. I know other proofs, for example the ratio test, may suffice in my example. I am just trying to shed light on my theoretical misunderstandings.
Thanks in advance.
 A: It is true that $\lim_{n\to \infty} |a_n| = 0 \implies \lim_{n\to \infty} a_n =0$. And the reverse implication is also true! This is the more obvious direction. Since $|x|$ is a continuous function, then $\lim_{n\to \infty} a_n = L \implies \lim_{n\to\infty} |a_n| = |L|$ for any convergent sequence $a_n$.
This applies to your series because $\lim_{n\to\infty} a_n = 0$ is a necessary condition for $\sum_n a_n$ to converge, even conditionally (and thus absolutely as well).
A: Let $s_n = \sum_{i=0}^n a_i$ be a convergent series, i.e. $\lim_{n \rightarrow \infty} s_n = L$ for some $L \in \mathbb{R}$. Then that means for every $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that $n \geq N \implies |s_n - L| < \epsilon$.
Now if we choose $N$ such that $n \geq N \implies |s_n - L| < \frac{\epsilon}{2}$, and choose $n$ such that $n - 1 \geq N$, we can write:
$\begin{eqnarray}|a_n| & = & |s_n - s_{n-1}| \\
& = & |(s_n - L) + (L - s_{n-1})| \\
& \leq & |s_n - L| + |L - s_{n-1}| & \textrm{by the triangle inequality} \\
& = & |s_n - L| + |s_{n-1} - L| & \textrm{since } |x| = |-x| \\
& < & \frac{\epsilon}{2} + \frac{\epsilon}{2} \\
& = & \epsilon \end{eqnarray}$
and hence $\lim_{n \rightarrow \infty} |a_n| = 0$.
So any convergent series requires that the terms of the sequence being summed to converge to zero.
