# find examples of two series $\sum a_n$ and $\sum b_n$ both of which diverge but for which $\sum \min(a_n, b_n)$ converges

Find examples of two series $\sum a_n$ and $\sum b_n$ both of which diverge but for which $\sum \min(a_n, b_n)$ converges. To make it more challenging, produce examples where $a_n$ and $b_n$ are positive and decreasing.

Edit: This problem is taken verbatim from Exercise 2.7.11 on page 68 of Abbott's Understanding analysis.

• So what have you tried? What are your favourite divergent series? What are your favourite convergent series? We will need a bit of both, won't we. Also, what are your favourite divergent series with positive decreasing terms? – Alex B. Dec 4 '10 at 14:15
• I have tried to start with (1/n) by replacing some terms with (1/(n^2)), but it dose not work. – chyojn Dec 4 '10 at 14:26
• That's a good start and we might need this later. Meanwhile, I have given you a hint for the first part in my answer below. – Alex B. Dec 4 '10 at 14:31
• Perhaps it is just a different edition, but Exercise 2.7.11 is on page 37 in the edition, which I was able to preview using Google Books. – Martin Sleziak Dec 5 '13 at 10:37

Here is a hint for the first part: you can make min$(a_n,b_n)=0$ for all $n$ (it doesn't get more convergent than that), while the two series $\sum a_n$ and $\sum b_n$ are very far from converging. Once you solve this one, let me know and I will give you a hint for the challenging part. It will build on ideas from the first one.

Edit: Nice solutions to the first part! Now for the second: of course the idea will have to be the same: the sequences will have to somehow alternate in which one is lower at any given point. But think about it: if they alternate at each step, like before, then we will have that $\sum_n a_{2n}$ converges and also $a_{2n+1} < a_{2n}$, so we will have that $\sum_n a_n < 2\sum_n a_{2n}$ will converge. That's not good. If you think about this issue for a while, you will realise that the intervals at which one sequence dives below the other one must get longer and longer for the two to diverge. Now, can you make your previous idea work?

• $a_n=0$, $b_n=1$, if $n$ is odd; $a_n = 1$, $b_n = 0$, if $n$ is even. – chyojn Dec 4 '10 at 14:35
• I am considering $min(a_n, b_n)=1/n^2$, and make the length of $n$th alternative interval equal to $n$, the value in the interval is $1/n^2$, then the sum in the interval will be $1/n$, and it seems to diverge ... – chyojn Dec 4 '10 at 15:24
• Very good, you got the basic idea now! You just have to be a bit careful and write out all the details of what the values of both $a_n$ and $b_n$ have to be to make it work. But the idea is exactly right: you want some partial sums of both $a_n$ and $b_n$ to contain $1/n$ and you don't care how long it takes to accumulate that $1/n$, i.e. how many terms you have to sum for each $1/n$. At the same time, you must make sure that the min really does converge. I am sure that you will be able to work out all the details from here. – Alex B. Dec 4 '10 at 15:41
• I see. Thank you very much. – chyojn Dec 4 '10 at 15:55

This answer gives a specific example using the ideas proposed in the previous answer:

Let $\sum a_n=\overline{\frac{1}{1^2}}+\frac{1}{2^2}+\overline{\frac{1}{3^2}+\frac{1}{3^2}+\frac{1}{3^2}}+\frac{1}{6^2}+\frac{1}{7^2}+\cdots+\frac{1}{14^2}+\overline{\frac{1}{15^2}+\frac{1}{15^2}+\cdots+\frac{1}{15^2}}+\cdots$

and $\sum b_n=\frac{1}{1^2}+\underline{\frac{1}{2^2}}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\underline{\frac{1}{6^2}+\frac{1}{6^2}+\cdots+\frac{1}{6^2}}+\frac{1}{15^2}+\frac{1}{16^2}+\cdots+\frac{1}{71^2}+\cdots$.

Notice that each group of repeated terms has a sum $S\ge\frac{1}{4}$, and that $\sum c_n=\sum\frac{1}{n^2}$.