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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.

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    $\begingroup$ 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? $\endgroup$
    – Alex B.
    Dec 4, 2010 at 14:15
  • $\begingroup$ I have tried to start with (1/n) by replacing some terms with (1/(n^2)), but it dose not work. $\endgroup$
    – chyojn
    Dec 4, 2010 at 14:26
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    $\begingroup$ 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. $\endgroup$
    – Alex B.
    Dec 4, 2010 at 14:31
  • $\begingroup$ 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. $\endgroup$ Dec 5, 2013 at 10:37
  • $\begingroup$ To me, it will never cease to be short of unbelievable how such a poorly elaborated question got 19 upvotes and 9 bookmarks. The question is not even original, as the OP admitted. Then I have to put up with people who say that questions should receive the same points per upvote than answers. $\endgroup$
    – Laz
    Sep 12, 2021 at 5:33

3 Answers 3

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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?

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    $\begingroup$ $a_n=0$, $b_n=1$, if $n$ is odd; $a_n = 1$, $b_n = 0$, if $n$ is even. $\endgroup$
    – chyojn
    Dec 4, 2010 at 14:35
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    $\begingroup$ 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 ... $\endgroup$
    – chyojn
    Dec 4, 2010 at 15:24
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    $\begingroup$ 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. $\endgroup$
    – Alex B.
    Dec 4, 2010 at 15:41
  • $\begingroup$ I see. Thank you very much. $\endgroup$
    – chyojn
    Dec 4, 2010 at 15:55
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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 $\sum c_n=\sum\frac{1}{n^2}$.

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Preamble

For the purposes of this answer, a substring of a sequence $(x_n)$ will mean a finite sequence of the form $(x_n)_{n=A}^B$.


Lemma 1: For all $\epsilon, L \in (0,\infty)$, there is a substring of $(\frac{1}{n})$ whose first term is less than $\epsilon$ and whose sum exceeds or equals L.

Proof: The Archimedean property guarantees that there is some $N \in \mathbf{N}$ such that $\frac{1}{N} < \epsilon$. $\sum_{n=1}^\infty \frac{1}{N+n-1}$ diverges, so there is $M \in \mathbf{N}$ such that $\sum_{n=N}^{N+M-1} \frac{1}{n} = \sum_{n=1}^{M} \frac{1}{N+n-1} \geq L$. Then $(\frac{1}{n})_{n=N}^{N+M-1}$ is a substring of $(\frac{1}{n})$ whose first term is less than $\epsilon$ and whose sum exceeds or equals L.


Lemma 2: For all $\epsilon \in (0,\infty)$ and $l \in \mathbf{N}$, there is a substring of $(\frac{1}{n^2})$ whose first term is less than $\epsilon$ and whose length equals $l$.

Proof: The Archimedean property guarantees that there is $N \in \mathbf{N}$ such that $\frac{1}{N} < \sqrt{\epsilon} \iff \frac{1}{N^2} < \epsilon$. Then $(\frac{1}{n^2})_{n=N}^{N+l-1}$ is a substring of $(\frac{1}{n^2})$ whose first term is less than $\epsilon$ and whose length equals $l$.


Construction

Construct infinite sequences of real numbers $(a_n)$ and $(b_n)$, as well as an infinite sequence of natural numbers $(N_k)$ as follows.

Base case: Let $a_1 = b_1 = N_1 = 1$.

Recursive step: For $k \in \mathbf{N}$ with $k>1$, there are two cases: $k$ is even and $k$ is odd.
[Case I: $k$ is even] Choose $N_k > N_{k-1}$ and $a_{N_{k-1}+1}, \ldots, a_{N_k}$ so that

  • $(a_n)_{n=N_{k-1}+1}^{N_k}$ is equal to a substring of $(\frac{1}{n})$
  • $a_{N_{k-1}+1} < a_{N_{k-1}}$
  • $\sum_{n=N_{k-1}+1}^{N_k} a_n \geq 1$.

Lemma 1 guarantees that such an $N_k$ and substring of $(\frac{1}{n})$ exist to satisfy these conditions.
Choose $b_{N_{k-1}+1}, \ldots, b_{N_k}$ so that

  • $(b_n)_{n=N_{k-1}+1}^{N_k}$ is equal to a substring of $(\frac{1}{n^2})$.
  • $b_{N_{k-1}+1} < \min \{ b_{N_{k-1}}, a_{N_{k-1}} \}$.

Lemma 2 guarantees that such a substring of $(\frac{1}{n^2})$ exists to satisfy these conditions.

[Case II: $k$ is odd] Choose $N_k > N_{k-1}$ and $b_{N_{k-1}+1}, \ldots, b_{N_k}$ so that

  • $(b_n)_{n=N_{k-1}+1}^{N_k}$ is equal to a substring of $(\frac{1}{n})$
  • $b_{N_{k-1}+1} < b_{N_{k-1}}$
  • $\sum_{n=N_{k-1}+1}^{N_k} b_n \geq 1$.

Lemma 1 guarantees that such an $N_k$ and substring of $(\frac{1}{n})$ exist to satisfy these conditions.
Choose $a_{N_{k-1}+1}, \ldots, a_{N_k}$ so that

  • $(a_n)_{n=N_{k-1}+1}^{N_k}$ is equal to a substring of $(\frac{1}{n^2})$.
  • $a_{N_{k-1}+1} < \min \{ a_{N_{k-1}}, b_{N_{k-1}} \}$.

Lemma 2 guarantees that such a substring of $(\frac{1}{n^2})$ exists to satisfy these conditions.

The base case specifies $a_1$, $b_1$, and $N_1$. For $k>1$, the recursive step specifies $N_k > N_{k-1}$, $a_{N_{k-1}+1}, \ldots, a_{N_k}$, and $b_{N_{k-1}+1}, \ldots, b_{N_k}$. So each of $(a_n)$, $(b_n)$, and $(N_k)$ are well defined.


For all $n \in \mathbf{N}$ with $n > 1$, there are two cases: $n = N_{k-1}+1$ for some $k \in \mathbf{N}$ and $N_{k-1}+1 < n \leq N_k$ for some $k \in \mathbf{N}$.
[Case I: $n = N_{k-1}+1$ for some $k \in \mathbf{N}$] Now there are two subcases: $k$ is even and $k$ is odd. [Subcase I: $k$ is even] $a_n = a_{N_{k-1}+1} < a_{N_{k-1}} = a_{n-1}$ and $b_n = b_{N_{k-1}+1} < \min \{ b_{N_{k-1}}, a_{N_{k-1}} \} \leq b_{N_{k-1}} = b_{n-1}$. [Subcase II: $k$ is odd] $a_n = a_{N_{k-1}+1} < \min \{ a_{N_{k-1}}, b_{N_{k-1}} \} \leq a_{N_{k-1}} = a_{n-1}$ and $b_n = b_{N_{k-1}+1} < b_{N_{k-1}} = b_{n-1}$.
[Case II: $N_{k-1}+1 < n \leq N_k$ for some $k \in \mathbf{N}$] $(a_n)_{n=N_{k-1}+1}^{N_k}$ and $(b_n)_{n=N_{k-1}+1}^{N_k}$ are both equal to substrings of decreasing sequences, so $a_n < a_{n-1}$ and $b_n < b_{n-1}$. Since $a_n < a_{n-1}$ and $b_n < b_{n-1}$ in both cases, $\mathbf{(a_n)}$ and $\mathbf{(b_n)}$ are decreasing.

Also note that, aside from the first (positive) term of each sequence, $(a_n)$ and $(b_n)$ are made of terms from positive sequences, so $\mathbf{(a_n)}$ and $\mathbf{(b_n)}$ are positive.


Divergence Proof

For all $k \in \mathbf{N}$, $$ \sum_{n=1}^{N_{2k}} a_n = (a_{N_1} + \sum_{n=N_1+1}^{N_2} a_n) + \ldots + (\sum_{n=N_{2k-2}+1}^{N_{2k-1}} a_n + \sum_{n=N_{2k-1}+1}^{N_{2k}} a_n) $$

$$ > (\sum_{n=N_1+1}^{N_2} a_n) + \ldots + (\sum_{n=N_{2k-1}+1}^{N_{2k}} a_n) \geq 1 + \ldots + 1 = k $$

and

$$ \sum_{n=1}^{N_{2k}} b_n = (b_{N_1} + \sum_{n=N_1+1}^{N_2} b_n) + \ldots + (\sum_{n=N_{2k-2}+1}^{N_{2k-1}} b_n + \sum_{n=N_{2k-1}+1}^{N_{2k}} b_n) $$

$$ > (b_{N_1}) + \ldots + (\sum_{n=N_{2k-2}+1}^{N_{2k-1}} b_n) \geq 1 + \ldots + 1 = k. $$

So the order limit theorem implies $(\sum_{n=1}^{N_{2k}} a_n)$ and $(\sum_{n=1}^{N_{2k}} b_n)$ diverge. Consequently, Theorem 2.5.2 from Abbott (subsequences of convergent sequences converge to the same value) implies $\sum a_n$ and $\sum b_n$ both diverge.


Convergence Proof

Define a sequence of real numbers $(c_n)$ as follows. Let $c_1 = a_1$. For $n \in \mathbf{N}$ with $n>1$, let $c_n = b_n$ if $N_{k-1} < n \leq N_k$ for some even $k$, otherwise let $c_n = a_n$ (when $N_{k-1} < n \leq N_k$ for some odd $k$).

Claim: $c_n \leq \frac{1}{n^2}$. The claim is true for $n=1$ since $c_1 = 1 = \frac{1}{1^2}$. Suppose it is true for $n=m-1$. There are two cases: $m = N_{k-1}+1$ for some $k \in \mathbf{N}$ and $N_{k-1}+1 < m \leq N_k$ for some $k \in \mathbf{N}$.
[Case I: $m = N_{k-1}+1$ for some $k \in \mathbf{N}$] Now there are two subcases: $k$ is even and $k$ is odd. [Subcase I: $k$ is even] $c_m = c_{N_{k-1}+1} = b_{N_{k-1}+1} < \min \{ b_{N_{k-1}}, a_{N_{k-1}} \} \leq a_{N_{k-1}} = c_{N_{k-1}} = c_{m-1} \leq \frac{1}{(m-1)^2}$ and $b_{N_{k-1}+1} \in \{ \frac{1}{n^2} \mid n \in \mathbf{N} \}$, so $c_m = b_{N_{k-1}+1} \leq \frac{1}{m^2}$. [Subcase II: $k$ is odd] $c_m = c_{N_{k-1}+1} = a_{N_{k-1}+1} < \min \{ a_{N_{k-1}}, b_{N_{k-1}} \} \leq b_{N_{k-1}} = c_{N_{k-1}} = c_{m-1} \leq \frac{1}{(m-1)^2}$ and $a_{N_{k-1}+1} \in \{ \frac{1}{n^2} \mid n \in \mathbf{N} \}$, so $c_m = a_{N_{k-1}+1} \leq \frac{1}{m^2}$.
[Case II: $N_{k-1}+1 < m \leq N_k$ for some $k \in \mathbf{N}$] Again there are two subcases: $k$ is even and $k$ is odd. [Subcase I: $k$ is even] $c_m = b_m < b_{m-1} = c_{m-1} \leq \frac{1}{(m-1)^2}$ and $b_m \in \{ \frac{1}{n^2} \mid n \in \mathbf{N} \}$, so $c_m = b_m \leq \frac{1}{m^2}$. [Subcase II: $k$ is odd] $c_m = a_m < a_{m-1} = c_{m-1} \leq \frac{1}{(m-1)^2}$ and $a_m \in \{ \frac{1}{n^2} \mid n \in \mathbf{N} \}$, so $c_m = a_m \leq \frac{1}{m^2}$.
Since $c_n < \frac{1}{n^2}$ is true for $n=m$ in all cases, it must be true for all $n \in \mathbf{N}$.

For all $n \in \mathbf{N}$, one of $a_n$ or $b_n$ equals $c_n$, so $\min \{ a_n, b_n \} \leq c_n \leq \frac{1}{n^2}$. Since $\sum \frac{1}{n^2}$ converges, the comparison test implies $\sum \min \{ a_n, b_n \}$ converges.


Implementation

The following Haskell code implements this construction:

data Parity = Even | Odd

coupledSeq :: RealFrac a => (a,a) -> [(a,a)]
coupledSeq t = t : rec Even t
  where
    pair x y = (x,y)
    rec Even (an, bn) = list3 ++ rec Odd (last list3)
      where
        list1 = dsi an 1
        list2 = dssi (min bn an) (length list1)
        list3 = zipWith pair list1 list2
    rec Odd (an, bn) = list3 ++ rec Even (last list3)
      where
        list1 = dssi (min an bn) (length list2)
        list2 = dsi bn 1
        list3 = zipWith pair list1 list2

-- Descreasing Sequence of Inverses
--   Returns a substring of (1/n) whose first entry is less than
--   a given value and whose sum exceeds or equals another given
--   value. Analogous to Lemma 1.
dsi :: RealFrac a => a -> a -> [a]
dsi tol exc
  | tol <= 0  = error "First argument to dsi must be positive."
  | exc <= 0  = error "Second argument to dsi must be positive."
  | otherwise = map snd $ takeUntil ((>=exc) . fst) $ scanl1 acc
                [dupl $ (1/) $ fromInteger n | n <- [minInt..]]
  where
    takeUntil pred xs = (\ (x,y) -> x ++ [head y] ) $ span (not . pred) xs
    acc (a,_) (_,x) = (a+x,x)
    dupl x = (x,x)
    minInt = toInteger $ (+1) $ floor $ (1/) tol

-- Decreasing Sequence of Square Inverses
--   Returns a substring of (1/n^2) whose first entry is less than
--   a given value and whose length equals another given value.
--   Analogous to Lemma 2.
dssi :: RealFrac a => a -> Int -> [a]
dssi tol len
  | tol <= 0  = error "First argument to dssi must be positive"
  | len  < 1  = error "Second argument to dssi must be >=1"
  | otherwise = take len [(1/) $ fromInteger $ n*n | n <- [minInt..]]
  where
    minInt = toInteger $ ceiling $ sqrt $ fromInteger $ (+1) $ floor $ (1/) tol

We can use this to inspect the behavior of the two sequences. For example, by graphing $n^2 a_n$ and $n^2 b_n$ against $n$, we see the alternating behavior predicted by Alex B. . Both sequences "take turns being the bottom", and the intervals at which one sequence dives below the other one get longer and longer.

Plot of nna_n and nnb_n from n=1 to 10 Plot of nna_n and nnb_n from n=1 to 100 Plot of nna_n and nnb_n from n=1 to 1000 Plot of nna_n and nnb_n from n=1 to 6244143

The first few values of $N_k$ corresponding to this program are $N_1=1$, $N_2=4$, $N_3=33$, $N_4=1906$, $N_5=6244143$.

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