Two divergent series whose minimum converges?

Click For Summary

Discussion Overview

The discussion revolves around the challenge of identifying two divergent series, \(\Sigma a_n\) and \(\Sigma b_n\), such that the series formed by their minimum terms, \(\Sigma \min(a_n, b_n)\), converges. The participants emphasize that both sequences must be positive and decreasing.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant proposes a specific pair of sequences where the reciprocals are defined as \(a_n = 1, 4, 4, 16, 25, 25, 25, 25, 25, 25, 25, 144, 169, 196, \ldots\) and \(b_n = 1, 1, 9, 9, 9, 36, 49, 64, 81, 100, 121, 121, 121, 121, \ldots\), claiming that the series of minimum terms converges to the square harmonic series.
  • Another participant shares an initial attempt at constructing sequences, suggesting a_n and b_n based on a pattern of sums but ultimately finds that their minimum terms yield a divergent series instead.
  • A different suggestion involves defining \(a_n = 1/n\) and \(b_n = 1\) for odd \(n\) and \(b_n = -1/n\) for even \(n\), although this does not adhere to the positive and decreasing requirement.
  • A participant points out that the stipulation of both sequences being positive and decreasing was not initially clear and requests clarification on this point.

Areas of Agreement / Disagreement

Participants express differing approaches and models for constructing the sequences, with no consensus reached on a definitive solution. The discussion remains unresolved regarding the existence of such series that meet all stipulated conditions.

Contextual Notes

Some sequences proposed do not meet the requirement of being positive and decreasing, highlighting the importance of adhering to the problem's constraints. Additionally, the complexity of defining sequences algorithmically versus finding closed-form expressions is noted.

coffeetheorem
Messages
3
Reaction score
0
I am struggling to find two divergent series, \Sigmaan and \Sigmabn, such that the series of minimum terms, \Sigmamin{an,bn}, actually converges.

A further stipulation is that both an and bn must be positive, decreasing sequences. (Otherwise the problem is trivial, as one could simply alternate 1/n and 1/n^2 to achieve the desired result.)
 
Last edited:
Physics news on Phys.org
Here's one way. For concision I'm just going to list the reciprocals of the terms:

Code:
{a_n} = 1  4  4 16  25  25  25  25  25  25   25 144 169 196 ...

{b_n} = 1  1  9  9   9  36  49  64  81 100  121 121 121 121 ...

Clearly these are positive and decreasing, and the series of minimum terms is the square harmonic series, which converges. But we let the two series take increasingly long "turns" being constant, long enough that each can be seen to be at least as large as the harmonic series.

For instance, the bottom series needs to stay long enough at 1/121 so that the part after 9,9,9 (which corresponds to 1/3) sums to at least (1/4).
 
Thanks, boss!

I had attempted a pair that was similar, but too simple and not quite workable. Here's what I initially tried (in reciprocal):

a_n = 1+9+9+ 9+ 25+25+25+25+25+49+49+49+49+49+49+49+81 … (nine times) …
b_n = 4+4+16+16+16+16+36+36+36+36+36+36+64 … (eight times) …

Then a_n is (2k+1)/(2k+1)^2 = 1/(2k+1), b_n is 2k/(2k) ^2 = 1/2k,

but the sequence of minimum terms is [k/2]/k^2 (where [k/2] denotes the floor function), which is approximately 1/2k.

Your sequences seem to be neatly-defined, letting each "go out far enough" to get the next term in the (basic) harmonic series. I would be hard pressed to find a closed-form expression for them, although I'd think it would be easy to define them algorithmically.
 
Another would be to let a_n=1/n, b_n=1 if n odd, b_n=-1/n if n even.
 
arildno: Part of the stipulation of the problem is that both sequences must be positive and decreasing. I have edited the original post to clarify this.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 17 ·
Replies
17
Views
6K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
6K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 6 ·
Replies
6
Views
4K
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K