- #1

Nick89

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## 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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