Decomposing a Series Using Riemann's Rearrangement Theorem

In summary, the conversation discusses the computation of the sum of a given series through the use of Riemann's Rearrangement Theorem. The series is split into two sums, one consisting of every odd indexed term and the other consisting of every even indexed term. The solution involves finding the sum of each individual series and then combining them to get the final sum.
  • #1
the_kid
116
0

Homework Statement


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

S=1+[itex]\frac{1}{4}[/itex]-[itex]\frac{1}{16}[/itex]-[itex]\frac{1}{64}[/itex]+[itex]\frac{1}{256}[/itex]


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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  • #2
the_kid said:

Homework Statement


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

S=1+[itex]\frac{1}{4}[/itex]-[itex]\frac{1}{16}[/itex]-[itex]\frac{1}{64}[/itex]+[itex]\frac{1}{256}[/itex]


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
 
  • #3
My apologies; I should have been clearer in my original post.

The signs are ++-- ++-- ...
 
  • #4
Any help?
 
  • #5
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 = [itex]\sum[/itex][itex](-1/16)[/itex]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!
 

FAQ: Decomposing a Series Using Riemann's Rearrangement Theorem

What is the Riemann Rearrangement Theorem?

The Riemann Rearrangement Theorem is a mathematical theorem that states that given an infinite series of real numbers, the terms of the series can be rearranged in any way and still converge to the same limit, as long as the series is conditionally convergent.

What does it mean for a series to be conditionally convergent?

A series is said to be conditionally convergent if it converges, but not absolutely. This means that while the series itself converges, the absolute value of the terms in the series do not converge.

Can the Riemann Rearrangement Theorem be applied to all infinite series?

No, the Riemann Rearrangement Theorem can only be applied to conditionally convergent series. For absolutely convergent series, the terms must be rearranged in a specific order in order to maintain the same limit.

How was the Riemann Rearrangement Theorem proven?

The Riemann Rearrangement Theorem was first proven by German mathematician Bernhard Riemann in the mid-19th century. His proof involved using a technique called the Dirichlet test, which is commonly used to test for the convergence of infinite series.

What are some real-world applications of the Riemann Rearrangement Theorem?

The Riemann Rearrangement Theorem has applications in various fields such as physics, engineering, and economics. It is used to analyze and model complex systems that involve infinite series, such as fluid dynamics and signal processing. It is also used in financial forecasting and risk management.

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