Two divergent series whose minimum converges?

In summary, it is possible to find two divergent series, \Sigmaan and \Sigmabn, that converge when their minimum terms, \Sigmamin{an,bn}, are taken. To meet the stipulation that both an and bn must be positive and decreasing, one approach is to alternate between long sequences of constant terms in each series, ensuring that each series is at least as large as the harmonic series. Another approach is to define the series algorithmically, such as letting a_n=1/n and b_n=1 if n is odd, and b_n=-1/n if n is even.
  • #1
coffeetheorem
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I am struggling to find two divergent series, [tex]\Sigma[/tex]an and [tex]\Sigma[/tex]bn, such that the series of minimum terms, [tex]\Sigma[/tex]min{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.)
 
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  • #2
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).
 
  • #3
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.
 
  • #4
Another would be to let a_n=1/n, b_n=1 if n odd, b_n=-1/n if n even.
 
  • #5
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.
 

1. What is a divergent series?

A divergent series is a sequence of numbers that does not have a finite limit as the number of terms approaches infinity. In other words, the sum of the terms in the series does not converge to a single value.

2. What is a convergent series?

A convergent series is a sequence of numbers that has a finite limit as the number of terms approaches infinity. In other words, the sum of the terms in the series converges to a single value.

3. Can two divergent series have a minimum that converges?

Yes, it is possible for two divergent series to have a minimum that converges. This means that although the two series individually do not converge, the minimum of the two series does converge.

4. How can two divergent series have a minimum that converges?

This can happen when the two series have terms that approach infinity at different rates. For example, one series may have terms that increase at a slower rate, while the other series has terms that increase at a faster rate. As a result, the minimum of the two series may have terms that increase at a rate that allows for convergence.

5. Is there a mathematical explanation for two divergent series having a minimum that converges?

Yes, there is a mathematical explanation for this phenomenon. It involves concepts such as limits, sequences, and series. By analyzing the behavior of the two series and their terms, it is possible to determine whether their minimum will converge or not.

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