# Alternating combination of a convergent and a divergent series

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.
• My problem is that I thought the rearrangement of $\sum_{n=1}^k s_n = \sum_{n=1}^k a_n -\sum_{n=1}^k b_n$ is not valid as you take it going to infinity, as we can’t rearrange the sum without knowing that it is absolutely convergent? Is this not correct? Commented Apr 14, 2018 at 21:06