I have been struggling with this question for days now and I feel that I must be missing something simple. I can see intuitively why the result is true but I can't come up with a proof. The question is as follows:
Let $a_n$ and $b_n$ be non negative numbers such that $\sum_{n=1}^\infty a_n$ diverges and $\sum_{n=1}^\infty b_n$ converges. Define a series $(s_n)_n$ by setting $s_1 = a_1$, $s_2 = a_1-b_1$ and $s_3 = a_1-b_1+a_2$ and so on for general $n\geq 2$:
$s_{2n-1} = a_1 - b_1 + a_2 - b_2 + ... + (a_{n-1}-b_{n-1}) + a_n$
$s_{2n} = a_1 - b_1 + a_2 - b_2 + ... + (a_n - b_n)$
Show that $(s_n)_n$ diverges.
I am struggling to make any use of the information about the $a_n$ and $b_n$ series. Everything I can think of involves rearranging the sum but as I understand it I can't do this since the rearrangement is not necessarily valid as the sums go to $\infty$.
The only attempt I came up with was to focus on the odd and even subsequences seperately. I found that for the even subsequence the difference between terms is $a_n - b_n$ and for the odd subsequence $a_{n+1}-b_n$. I know that from the first comparison test $a_n$ must be greater than $b_n$ for infinitely many terms or $\sum_{n=1}^\infty a_n$ would be convergent. As far as I can see this means there are infinitely many times when the difference between terms is positive, but there could still be more times where it is negative so I can't see how to get from this that the sums are always growing.
I would appreciate if anyone could give me some sort of hint of the right direction to go in.
Call $A = \sum_{n = 1}^\infty A_n$. The intuition here is that $(s_n)_{n = 1}^\infty$ is basically tending towards the limit $A$, minus the $b_n$ series. That is, past some point, the added $a_n$ terms to the sequence $\sum s_n$ won't really add anything; and then we will be left with just the $b_n$, which since they diverge, will cause the whole $\sum s_n$ to diverge.
So, to show that $\sum_{n =1 }^\infty s_n$ diverges, pick some number $M$. For some large $N$, $|A - \sum_{n = 1}^k a_n|$ is very small for all $k \geq N$. Thus, $\sum_{n = 1}^k s_n = \sum_{n = 1}^k a_n - \sum_{n = 1}^k b_n$ is very nearly $A - \sum_{n = 1}^k b_n$. But the $b_n$ series is going off to infinity.