Sum of Series w/ 2's & 3's: Find the Total

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Homework Help Overview

The discussion revolves around finding the sum of a series composed of the reciprocals of positive integers whose only prime factors are 2's and 3's. Participants explore various representations of the series, including attempts to express it in sigma notation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss breaking down the series into geometric series and consider the implications of writing it in sigma notation. There are inquiries about the general form of numbers with specific prime factors and whether the sum can be expressed as a double sum.

Discussion Status

There is an ongoing exploration of how to represent the series mathematically, with some participants suggesting that sigma notation may clarify the problem. Others note that while the double sum is technically different from the original sum, it remains absolutely convergent. The conversation includes hints towards finding alternative methods of representation.

Contextual Notes

Participants express uncertainty about the calculations and representations, indicating a lack of consensus on the best approach to take. There are references to technical notes regarding the convergence of the series and the potential for rearranging terms.

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find the sum of the series 1 + 1/2 + 1/3 + 1/4 + 1/6 + 1/8 + 1/9 + 1/12 + ...,
where the terms are the reciprocals of the positive integers whose only prime factors are 2's and 3's.

Here's my work so far:

1 + 1/2 + 1/3 + 1/4 + 1/6 + 1/8 + 1/9 + 1/12 + ... =
1 + 1/2 + 1/4 + 1/8 + 1/16 + ... + 1/3 + 1/9 + 1/27 + ... + 1/6 + 1/12 + 1/18 + 1/24 + ... =
1 + 1/2 + 1/4 + 1/8 + 1/16 + ... + 1 + 1/3 + 1/9 + 1/27 + ... + 1 + 1/6 + 1/12 + 1/18 + 1/24 + ... - 2

Those first two are geometric series. I don't know what to do with the third one.
 
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Try writing it in sigma notation. (Σ)
 
Hurkyl said:
Try writing it in sigma notation. (Σ)

:confused:
My problem is to find the value of 1 + 1/6 + 1/12 + 1/18 + 1/24 + 1/32 + 1/48 + 1/54...
 
That sum is no easier to calculate than the original sum. (In fact, I would be able to figure out the original sum faster than the one in your last post)

There are other ways to see how to approach this problem, but IMHO, writing it in sigma notation makes the right step much clearer.

In general, it's good practice to write sums in sigma notation anyways.
 
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Hurkyl said:
That sum is no easier to calculate than the original sum. (In fact, I would be able to figure out the original sum faster than the one in your last post)

There are other ways to see how to approach this problem, but IMHO, writing it in sigma notation makes the right step much clearer.

In general, it's good practice to write sums in sigma notation anyways.

Well then my problem is "sigma notation". :cry:
 
Do you know how to write a simpler sum in sigma notation? Like 1 + 1/2 + 1/4 + 1/8 + 1/16 + ...?
 
Hurkyl said:
Do you know how to write a simpler sum in sigma notation? Like 1 + 1/2 + 1/4 + 1/8 + 1/16 + ...?

Yes.
[tex]\sum_{n=0}^\infty \frac{1}{2^n}[/tex]
 
Ok, good. Next question: what's the general form of a number whose prime factors are only 2 and 3?
 
Hurkyl said:
Ok, good. Next question: what's the general form of a number whose prime factors are only 2 and 3?

That's [tex]2^n3^m[/tex]
 
  • #10
So does that suggest how to write the sum of the reciprocals of every positive integer whose prime factors can only be 2's and 3's in sigma notation?
 
  • #11
Hurkyl said:
So does that suggest how to write the sum of the reciprocals of every positive integer whose prime factors can only be 2's and 3's in sigma notation?

Indeed. Is it going to be a double sum?
 
  • #12
Yes, it is!


Technical note: actually, the double sum is technically different from the original sum. However, the sum turns out to be absolutely convergent (the easiest way to see this fact is that it has no negative terms!), and this sort of "rearrangement" is perfectly legal with absolutely convergent sums, meaning that they'll give the same answer.
 
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  • #13
Hurkyl said:
Yes, it is!

And is the correct answer 3? :blushing:
 
  • #14
Sounds right. Well done! By the way, don't miss the technical note I edited into my last post.

By the way, now that you know how the problem is solved, it may be instructive to try and figure out how to do it without sigma notation. Then again, it might not be instructive too, I don't know. :smile:
 
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  • #15
Hurkyl said:
Sounds right. Well done! By the way, don't miss the technical note I edited into my last post.

By the way, now that you know how the problem is solved, it may be instructive to try and figure out how to do it without sigma notation. Then again, it might not be instructive too, I don't know. :smile:

Thanks for the technical note. Had forgot it. o:)
Without sigma notation? I have no clue at all :-p
 
  • #16
You had it as a double sum right? With a bit of rearranging you should be able to write it as a product of 2 infinite sums, which I suppose you could write without sigma notation if you liked. I'd suggest you try this as the product form you'll get is quite cute and worth the effort.
 
  • #17
shmoe said:
You had it as a double sum right? With a bit of rearranging you should be able to write it as a product of 2 infinite sums, which I suppose you could write without sigma notation if you liked. I'd suggest you try this as the product form you'll get is quite cute and worth the effort.

Yes, that's exactly what I did.
 

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