Exercise 2.7.11

Question

Find examples of two series $\sum a_{n}$ and $\sum b_{n}$ both of which diverge but for which $\sum \min \left\{a_{n}, b_{n}\right\}$ converges. To make it more challenging, produce examples where $\left(a_{n}\right)$ and $\left(b_{n}\right)$ are strictly positive and decreasing.

Ulisse Mini · Accepted Answer

Let $m_n = \min\{a_n, b_n\}$.
  Clearly $(m_n)$ must take an infinite amount of $(a_n)$ and $(b_n)$ terms, as otherwise removing the finite terms would imply one of $\sum a_n$ or $\sum b_n$ converged.

The key insight is that as long as $a_n, b_n > 0$, we can simply repeat terms in one sequence (while letting $m_n$ be governed by the other sequence) for as long as we want - say, until we have enough terms to e.g. sum to 1. Then we can switch the roles of the sequences.
    To start with the construction, take some converging series with all terms positive - say, $m_n = 1/2^n$.
    Then, define the first few terms of $m_n$, $a_n$, and $b_n$ as:

$$
\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}
    \hline
n   &1 &2 &3 &[4, 4 + 8 = 12)& [12, 12 + 2^{11}) & [12 + 2^{11}, 12 + 2^{11} + 2^{12 + 2^{11}})\ \hline
m_n &1/2 &1/4 & 1/8 & 1/2^n & 1/2^n & 1/2^n \ \hline
\min\{a_n, b_n\} &a_n &b_n & b_n & a_n & b_n & a_n\ \hline
a_n &1/2 & 1/2 & 1/2 & 1/2^n & 1/2^{11} & 1/2^n\ \hline
b_n &1 & 1/4 & 1/8 & 1/8 & 1/2^n & 1/2^{12 + 2^{11}}\ \hline
\end{array}
$$

Specifically, $\min\{a_n, b_n\}$ will alternate between following $a_n$ and $b_n$. Every ``block'' of the sequence (examples: each of the last three columns) that isn't being followed by $\min\{a_n, b_n\}$ sums to 1.
Since each block is finite, $\min\{a_n, b_n\}$ will alternate between $a_n$ and $b_n$ infinitely, and thus both $\sum a_n$ and $\sum b_n$ will diverge.

For the sake of completeness, $a_n$ and $b_n$ are defined more formally below.
Let $k_1 = 1$, $k_n = k_{n-1} + 2^{k_{n-1} - 1}$. Then
$$
    a_n = \begin{cases}
        1/2^n & \text{ if } k_{2p - 1} \leq n < k_{2p} \
        1/2^{k_{2p} - 1} &\text{ if } k_{2p} \leq n < k_{2p + 1}
    \end{cases}
\text{ and }
    b_n = \begin{cases}
        1/2^n & \text{ if } k_{2p} \leq n < k_{2p + 1} \
        1/2^{k_{2p - 1} - 1} &\text{ if } k_{2p - 1} \leq n < k_{2p}
    \end{cases}
$$

% Preliminary results/requirements on a_n and b_n:
%   Let $m_n = \min\{a_n, b_n\}$.
%   Clearly $(m_n)$ must take an infinite amount of $(a_n)$ and $(b_n)$ terms, as otherwise removing the finite terms would imply one of $\sum a_n$ or $\sum b_n$ converged.

%   Because $(m_n) \to 0$ and because $(m_n)$ contains an infinite amount of $(a_n)$ terms we can find $N$ large enough that $a_N < \epsilon$, because $(a_n)$ is decreasing this implies every $n \ge N$ has $a_n < \epsilon$, which means $(a_n) \to 0$. The same logic shows $(b_n) \to 0$.

%   If we forget about the decreasing requirement we can make $a_n$ alternate between $1/n$ and $1/n^2$ and have $b_n$ do the same, then $m_n = 1/n^2$ converges.

%   A similar example won't work here, since if $\sum a_{2n-1}$ converges then $a_{2n} < a_{2n-1}$ implies $\sum a_{2n}$ converges by the comparison test, combining these shows $\sum a_n$ converges. Therefor $(m_n)$ can't alternate terms, and we need to be more clever.

%   In fact, If $\sum a_{n_k}$ converges then every shifted sum $\sum a_{n_k+p}$ will converge by a similar argument as above. Meaning $\left(a_{n_k}\right)$ can't be periodic since that would make $(a_n)$ converge.

% TODO: Same exercise for functions? add appendix with my own exercises

Exercise 2.7.11

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Exercise 2.7.11

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