Alternating series test question. I'm reading this in my text:

So the alternating series test says:

i) is about the sequence decreasing
ii) is about the limit of the b term going to 0
I'm confused about why we need to do anything more once we find out that ii) isn't satisfied in example 2. What does it mean that we're looking at the limit of the nth term of the series? Don't we know that it diverges already? Also, can someone show me how they determine the limit of $a_n$?

Here is a discussion
of a more general case of this.
Suppose
$b_n \ge b_{n+1}$
and
$\lim_{n \to \infty} b_n
=b > 0
$.
Let
$s_n
=\sum_{k=1}^n (-1)^nb_n
$.
How does $s_n$ behave?
Let
$t_n
=\sum_{k=1}^n (-1)^n(b_n-b)
$.
Since
$b_n-b \to 0$
and $b_n-b$
is decreasing,
$\lim_{n \to \infty} t_n$
exists.
Call it $T$.
$t_n
=\sum_{k=1}^n (-1)^kb_k+\sum_{k=1}^n (-1)^kb
=\sum_{k=1}^n (-1)^kb_k+b\sum_{k=1}^n (-1)^k
$
so
$\sum_{k=1}^n (-1)^kb_k
=t_n-b\sum_{k=1}^n (-1)^k
$.
Therefore
$\begin{array}\\
\sum_{k=1}^{2n} (-1)^kb_k
&=t_{2n}-b\sum_{k=1}^{2n} (-1)^k\\
&=t_{2n}\\
&\to T\\
\text{and}\\
\sum_{k=1}^{2n+1} (-1)^kb_k
&=t_{2n+1}-b\sum_{k=1}^{2n+1} (-1)^k\\
&=t_{2n+1}+b\\
&\to T+b\\
\end{array}
$
Note that
$\sum_{k=1}^{2n} (-1)^kb_k
=\sum_{k=1}^{n} (b_{2k}-b_{2k-1})
\to T
$.
Your case is
$b_n
=\dfrac{3n}{4n-1}
$
so, as you wrote
$b = \dfrac34
$
so the limit points are
$T$
and
$T+\dfrac34$
where
$\begin{array}\\
T
&=\lim_{n \to \infty} \sum_{k=1}^{n} (b_{2k}-b_{2k-1})\\
&=\lim_{n \to \infty} \sum_{k=1}^{n}(\dfrac{6k}{8k-1}-\dfrac{3(2k-1)}{4(2k-1)-1})\\
&=\lim_{n \to \infty} \sum_{k=1}^{n}(\dfrac{6k}{8k-1}-\dfrac{6k-3}{8k-5})\\
&=\lim_{n \to \infty} \sum_{k=1}^{n}\dfrac{6k(8k-5)-(6k-3)(8k-1)}{(8k-1)(8k-5)}\\
&=\lim_{n \to \infty} \sum_{k=1}^{n}\dfrac{48k^2-30k-(48k^2-30k+3)}{(8k-1)(8k-5)}\\
&=\lim_{n \to \infty} \sum_{k=1}^{n}\dfrac{-3}{(8k-1)(8k-5)}\\
&=\lim_{n \to \infty} -\dfrac{3}{64}\sum_{k=1}^{n}\dfrac{1}{(k-1/8)(k-5/8)}\\
\end{array}
$
Note that if
$b_n
=\dfrac{un}{vn+w}
$
then
$b = \dfrac{u}{v}
$
so the limit points are
$T$
and
$T+\dfrac{u}{v}$
where
$\begin{array}\\
T
&=\lim_{n \to \infty} \sum_{k=1}^{n} (b_{2k}-b_{2k-1})\\
&=\lim_{n \to \infty} \sum_{k=1}^{n}(\dfrac{u(2k)}{v(2k)+w}-\dfrac{u(2k-1)}{v(2k-1)+w})\\
&=\lim_{n \to \infty} \sum_{k=1}^{n}(\dfrac{2uk}{2vk+w}-\dfrac{2uk-u}{2vk-v+w})\\
&=\lim_{n \to \infty} \sum_{k=1}^{n}\dfrac{2uk(2vk-v+w)-(2uk-u)(2vk+w)}{(2vk+w)(2vk-v+w)}\\
&=\lim_{n \to \infty} \sum_{k=1}^{n}\dfrac{4uvk^2+2u(w-v)k-(4uvk^2+2k(uw-uv)-uw)}{(2vk+w)(2vk-v+w)}\\
&=\lim_{n \to \infty} \sum_{k=1}^{n}\dfrac{uw}{(2vk+w)(2vk-v+w)}\\
&=\lim_{n \to \infty} \dfrac{uw}{4v^2}\sum_{k=1}^{n}\dfrac{1}{(k+w/(2v))(k+(w-v)/(2v))}\\
&=\lim_{n \to \infty} \dfrac{uw}{4v^2}\sum_{k=1}^{n}\dfrac{1}{(k+w/(2v))(k+w/(2v)-1/2)}\\
\end{array}
$
