Showing alternating series is convergent by definition with condition $b_1 - b_2 < \sum_{n=1}^\infty (-1)^{n-1}b_n < b_1$ 
(i) Let $b_n>0$, $n\geq 1$, be decreasing such that $\lim_{n\to\infty}b_n=0$. Show by definition that $\sum_{n=1}^\infty (-1)^n b_n$ is convergent and $$b_1 - b_2 < \sum_{n=1}^\infty (-1)^{n-1}b_n < b_1.$$
  (ii) Let $b_n>0$ and for all $n\geq1$, and let $\lim_{n\to\infty}a_n=\infty$ such that 
  $$\lim_{n\to\infty} a_n\left(\dfrac{b_n}{b_{n+1}} - 1 \right) > 0.$$
  Show that $\sum_{n=1}^\infty (-1)^n b_n$ is convergent. 

I don't know how to do this problem, I've tried a few things but nothing even worth writing. I am studying for a qualifier, this is not homework. 
 A: Let $s_n:=\sum_{k=1}^n (-1)^{k-1}b_k$.
We write $s_{2n}=b_1+(b_3-b_2)+(b_5-b_4)+\cdots + (b_{2n-1}-b_{2n-2})-b_{2n}$ and note that the terms in parentheses are negative. Hence the sequence $\{s_{2n}\}_{n=1}^\infty$ is bounded above by $b_1$. Since $(b_{2k+1}-b_{2k+2})>0$ for each $k \in \mathbb{N}$ we also have
\begin{equation}s_{2n} < s_{2n}+( b_{2n+1}-b_{2n+2} )=s_{2(n+1)}.
\end{equation}
Therefore, $\{s_{2n}\}_{n=1}^\infty$ is a convergent sequence since it is bounded above and increasing.
Let $s:=\lim_{n \to \infty}s_{2n}$.
We write $s_{2n-1}=b_1-b_2+(b_3-b_4)+(b_5-b_6)+\cdots+(b_{2n-3}-b_{2n-2})+b_{2n-1}$ and note that the terms in parentheses are positive. Hence the sequence $\{s_{2n-1}\}_{n=1}^\infty$ is bounded below by $b_1-b_2$. Since $(b_{2k+1}-b_{2k})<0$ for each $k \in \mathbb{N}$ we also have
\begin{equation}s_{2n-1} > s_{2n-1}+( b_{2n+1}-b_{2n} )=s_{2n+1}.
\end{equation}
Therefore, $\{s_{2n-1}\}_{n=1}^\infty$ is a convergent sequence since it is bounded below and decreasing.
Since $s_{2n-1}=s_{2n}+b_{2n}$ and $\lim_{n \to \infty}b_{2n}=0$, it now also follows that $ \lim_{n \to \infty}s_{2n-1}=s$.
Note: The above argument is abridged as the propositions regarding how each subsequence is bounded rigorously follow by means of induction.
A: For the second question, the obvious thing to do is to try to reduce it to the first. From the condition, we see that $\frac{b_n}{b_{n+1}}>1$ except for finitely many $n$. So $(b_n)_{n\ge K}$ is a decreasing sequence for some $K$. This means that the nonnegative sequence $b_n$ converges to a limit $b$. If $b=0$, we can use part (i).
However, it is not true that $b=0$. Let $a_n=n^2$ and $b_n=1+\frac1n$. Then 
$$a_n\left(\frac{b_n}{b_{n+1}}-1\right)=\frac{n^2}{1+\frac1{n+1}}\left(\frac1n-\frac1{n+1}\right)=\frac{n^2}{n(n+2)}\to 1
$$
but $\sum_{n=1}^\infty (-1)^n b_n$ is not convergent, as $(-1)^nb_n$ does not tend to zero.
