# 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.

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 $$$$s_{2n} < s_{2n}+( b_{2n+1}-b_{2n+2} )=s_{2(n+1)}.$$$$ Therefore, $$\{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 $$$$s_{2n-1} > s_{2n-1}+( b_{2n+1}-b_{2n} )=s_{2n+1}.$$$$ Therefore, $$\{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.

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.