# Sum of a series.

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.

Hurkyl
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Gold Member
Try writing it in sigma notation. (&Sigma;)

Hurkyl said:
Try writing it in sigma notation. (Σ)

My problem is to find the value of 1 + 1/6 + 1/12 + 1/18 + 1/24 + 1/32 + 1/48 + 1/54...

Hurkyl
Staff Emeritus
Gold Member
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".

Hurkyl
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Gold Member
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.
$$\sum_{n=0}^\infty \frac{1}{2^n}$$

Hurkyl
Staff Emeritus
Gold Member
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 $$2^n3^m$$

Hurkyl
Staff Emeritus
Gold Member
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?

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?

Hurkyl
Staff Emeritus
Gold Member
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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Hurkyl said:
Yes, it is!

And is the correct answer 3?

Hurkyl
Staff Emeritus
Gold Member
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.

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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.

Thanks for the technical note. Had forgot it.
Without sigma notation? I have no clue at all :tongue:

shmoe