Decomposing a Series Using Riemann's Rearrangement Theorem

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SUMMARY

The discussion focuses on computing the sum of the series S=1+1/4-1/16-1/64+1/256 using Riemann's Rearrangement Theorem, which applies due to the series being absolutely convergent. Participants suggest decomposing the series into two separate sums: S1 for odd-indexed terms and S2 for even-indexed terms. The first sum, S1, is expressed as S1 = Σ(-1/16)^n from n = 0 to ∞. Further exploration of the second sum is encouraged, with a note that it requires a constant factor.

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Homework Statement


I'm trying to compute the sum of the following series:

S=1+\frac{1}{4}-\frac{1}{16}-\frac{1}{64}+\frac{1}{256}


Homework Equations





The Attempt at a Solution


I'm not really sure how to begin this one. I know it probably involves Riemann's Rearrangement Theorem since this series is absolutely convergent.
 
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the_kid said:

Homework Statement


I'm trying to compute the sum of the following series:

S=1+\frac{1}{4}-\frac{1}{16}-\frac{1}{64}+\frac{1}{256}


Homework Equations





The Attempt at a Solution


I'm not really sure how to begin this one. I know it probably involves Riemann's Rearrangement Theorem since this series is absolutely convergent.

Are the successive signs really ++--+? What happens after that? Are you sure you copied the question correctly?

RGV
 
My apologies; I should have been clearer in my original post.

The signs are ++-- ++-- ...
 
Any help?
 
Okay, so I think that if you think about it as the sum of two sums, that will help...

think of the first sum as the sum of every odd indexed term, and the second sum as the sum of every even indexed term.

S1 = 1-1/16+1/256...
S2 = 1/4-1/64+1/(16*64)...

thus, the first sum will be

S1 = \sum(-1/16)n from n = 0 to ∞.

and the second, I'll let you figure out. but I think that this should help. (note, the second one needs a constant out front.

hope this helps!
 

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