Stupid series I can't seem to find

In summary, the author is trying to find a function that will allow him to write the sum of the following series in a single sum. The sum should converge to 0.8, but they are not sure how to do it.
  • #1
Nick89
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0

Homework Statement



I need a way to write the following series in a sum:
[tex]\frac{1}{2} + \frac{1}{4} + \frac{1}{32} + \frac{1}{64} + \frac{1}{1024} + \frac{1}{2048} + ...[/tex]

If you look closely you can see a 2^n pattern in there:
[tex]\frac{1}{2} + \frac{1}{4} + \frac{0}{8} + \frac{0}{16} + \frac{1}{32} + \frac{1}{64} + \frac{0}{128} + \frac{0}{256} + \frac{1}{512} + \frac{1}{1024} + ...[/tex]

The denominator obviously follows [tex]2^n[/tex], but the numerator goes 110011001100...
I can't seem to find a function that will allow me to put this in a single sum:
[tex]\sum_{n=1}^{\infty} ...[/tex]

(The sum should converge to 0.8, right?)

The Attempt at a Solution


I tried using the mod operator to determine if n was even or odd, something like this:
[tex]\sum_{n=1}^{\infty} \frac{n \mod 2}{2^n} + \frac{n \mod 2}{2^{n+1}}[/tex]
This doesn't work, because the terms that are discarded when n is even are not discarded the next 'run' when (n+1) is even... Dunno how to explain this properly, but if you calculate it manually it does this:
[tex]\frac{1}{2} + \frac{1}{4} + \frac{0}{4} + \frac{0}{8} + \frac{1}{8} + \frac{1}{16} + ...[/tex]
So if you discard the 0/ ... terms you are just left with the usual 2^(-n) sum...

I have a feeling I'm close, but I can't figure it out :S
 
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  • #2
If the missing bits were not missing it would add up to 1.

Now you can put each missing bit into a relation with a bit that is there.

They are a certain fraction of terms that are there. So their sum is that proportion of 1 IYSWIM.

Do a diagram and this will look plausible.
 
  • #3
I can't seem to understand what you mean exactly...

By missing bits you mean the 0/x parts?
If so, then half of the terms as missing, so to your logic that would mean the sum would converge to 0.5...? (Whatr does IYSWIM mean?)

But I think it should converge to 0.8 (I said 8 before but that was a mistake, I edited it now).
 
  • #4
If it's the pattern you suggest, (1/2+1/4)*(1/16)=(1/32+1/64). (1/32+1/64)*(1/16)=(1/512+1/1024). It's a geometric series with a common ratio of 1/16. But this doesn't quite fit with the series you quote in the first line (the next term after 1/64 is 1/1024, not 1/512). Is that a typo?
 
  • #5
Nick89 said:
I can't seem to understand what you mean exactly...

By missing bits you mean the 0/x parts?
If so, then half of the terms as missing, so to your logic that would mean the sum would converge to 0.5...? (Whatr does IYSWIM mean?)

But I think it should converge to 0.8 (I said 8 before but that was a mistake, I edited it now).

IYSWIM means 'if you see what I mean'.
Just compare each pair of 'absent' terms with the 'present' pair preceding it.
 
  • #6
Dick said:
If it's the pattern you suggest, (1/2+1/4)*(1/16)=(1/32+1/64). (1/32+1/64)*(1/16)=(1/512+1/1024). It's a geometric series with a common ratio of 1/16. But this doesn't quite fit with the series you quote in the first line (the next term after 1/64 is 1/1024, not 1/512). Is that a typo?

It was indeed a typo. The correct form is right underneath, where 1/512 is present.

Alright, that makes a bit more sense. I still fail to see how I can now create a single sum for this series though... If I have a bit more time (maybe later tonight) I'll try to figure it out :)

Thanks so far.
 

1. What is the "Stupid series" about?

The "Stupid series" is a collection of books that satirize and critique various aspects of society, including politics, pop culture, and everyday life.

2. Who is the author of the "Stupid series"?

The author of the "Stupid series" is a pseudonymous writer known as "Stupidicus Maximus."

3. How many books are in the "Stupid series"?

As of now, there are 5 books in the "Stupid series," but the author has hinted at the possibility of future additions.

4. Is the "Stupid series" appropriate for all ages?

No, the "Stupid series" contains mature themes and language that may not be suitable for younger readers.

5. Where can I find the "Stupid series" if I can't seem to locate it?

The "Stupid series" is available for purchase on major online retailers such as Amazon and Barnes & Noble. It may also be available at local bookstores or libraries.

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