Find two divergent sequences whose minimum values create a convergent series | Step-by-Step Solution

What We're Solving

We need to construct an example where two divergent series have the special property that when we take the smaller value at each position, the resulting series actually converges! This is a fascinating example that shows how the interplay between sequences can create unexpected behavior.

The Approach

This problem is asking us to be creative mathematicians:

• We want two sequences that individually create divergent series
• But when we take their minimum values, most terms should be "small enough" to make the series converge
• The key insight is that we can make the sequences "take turns" being the smaller one

Think of it like two competing runners where each one occasionally falls behind the other, and when one falls behind, they fall really far behind!

Step-by-Step Solution

Step 1: Choose our first divergent sequence Let's start with a classic divergent series. We'll use: $$a_n = \frac{1}{n}$$

This gives us the harmonic series $\sum \frac{1}{n}$, which we know diverges. ✓ The sequence is decreasing and positive.

The Elegant Solution:

• $a_n = \frac{1}{n}$
• $b_n = \frac{1}{2^k}$ where $k = \lfloor \log_2 n \rfloor$

Step 3: Analyze what happens with the minimum For most values of $n$, we have $b_n = \frac{1}{2^k}$ where $k$ grows very slowly with $n$.

• When $n$ is small, sometimes $a_n < b_n$, sometimes $b_n < a_n$
• But the key insight is that $\min\{a_n, b_n\}$ behaves roughly like $\frac{1}{n^2}$ for large enough $n$

Step 4: Verify convergence The series $\sum \min\{a_n, b_n\}$ converges because it's dominated by a convergent $p$-series with $p > 1$.

The Answer

Here's a clean example:

• $a_n = \frac{1}{n}$ (harmonic series - diverges)
• $b_n = \begin{cases} \frac{1}{n^2} & \text{if } n = 2^k \text{ for some integer } k \\ \frac{2}{n} & \text{otherwise} \end{cases}$

The series $\sum b_n$ diverges (dominated by $\sum \frac{2}{n}$), but $\sum \min\{a_n, b_n\}$ converges because the minimum is $\frac{1}{n^2}$ at infinitely many points where $n = 2^k$, and these dominate the convergence.

Memory Tip

Think of this as the "teamwork paradox" - two sequences that individually fail to converge, but when they work together (taking the best of both), they succeed! The key is making sure one sequence occasionally becomes much smaller than the other at strategic points.

Great question! This type of problem really shows how creative and surprising mathematical analysis can be. Keep exploring these kinds of constructions - they build excellent mathematical intuition! 🌟