What does an infinte series =?

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Homework Statement



This isint really homework...but some kid at my school was telling me about the series 1/2+1/4+1/8+1/16+1/32+1/64...going on forever...now thinking about this it approaches 1...but apperantly is also infinte? could someone explain this? is this kid just full of himself?

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The Attempt at a Solution

 
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Suppose [itex]0 \leq r < 1[/itex] and fix [itex]k \in \mathbb{N}[/itex]. Then we have the following:
[tex](1-r)\sum_{n=0}^k r^n = \sum_{n=0}^k r^n - r\sum_{n=0}^k r^n = 1-r^{k+1}[/tex]
This allows us to evaluate the sum in the finite case. That is,
[tex]\sum_{n=0}^k r^n = \frac{1-r^{k+1}}{1-r}[/tex]
Evaluating the limit as [itex]k \rightarrow \infty[/itex] tells us how to evaluate the infinite sum. In particular, we have
[tex]\sum_{n=0}^{\infty} r^n = \frac{1}{1-r}[/tex]
The sum that you listed is the case when [itex]r = 2^{-1}[/itex]. This means we have
[tex]\sum_{n=0}^{\infty} 2^{-n} = 1 + \sum_{n=1}^{\infty} 2^{-n} = 2[/tex]
This tells you that
[tex]\sum_{n=1}^{\infty} 2^{-n} = 1[/tex]
So the sum converges to 1.
 
TheKracken said:

Homework Statement



This isint really homework...but some kid at my school was telling me about the series 1/2+1/4+1/8+1/16+1/32+1/64...going on forever...now thinking about this it approaches 1...but apperantly is also infinte? could someone explain this? is this kid just full of himself?

Homework Equations





The Attempt at a Solution


From the way this post is worded, I'm assuming you don't want to hear a lot of technical jargon. So I'll keep it simple.

Just in case you weren't aware of the nomenclature (because I'll make reference to that):

Sequence = ordering of terms.

Series = Sum of terms in a sequence.

This is called a geometric series. This is the sum of a sequence of terms where the next term is derived by multiplying the current term by a fixed number (called the common ratio). So by knowing the first term and the common ratio, you can get all the terms of the sequence, and by then summing them, find the sum of the series.

Here the first term is 1/2 and the common ratio is also 1/2. Meaning the terms progressively get smaller. They will never completely vanish (become zero), but they will become arbitrary close as you take more and more terms.

This is called an infinite series because you "let" the series grow to an "infinite" length i.e. you don't stipulate an end to it. Practically, you will never be able to write down all the terms, but the continuation (...) at the end means that you're theoretically considering an infinite number of terms. As the sequence goes on, the terms get closer to zero, and this is important for something known as convergence, which means that the sum actually exists and is finite. If the series doesn't converge, it just "blows up" to arbitrarily larger numbers (this is called diverging to infinity).

In this case, because of the common ratio of 1/2, the terms get smaller and smaller and the series converges. We can work out the sum of the infinite geometric series as jgens stated (here it's [itex]\frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1[/itex] as your friend said). This essentially means that if you take more and more terms and sum them up (i.e. do the "partial sums" for a greater and greater number of successive terms), you will get arbitrarily close to 1, but never quite hit it. Here the partial sums will always be strictly less than one because the terms are all positive.

Hope this has made things very clear to you.
 
think of it as taking a piece of paper of area 1, and then cutting this piece of paper in half then cut the remaining half in half again, and do this forever and then you will have an infinite number of rectangles and when you add them back up you get the original piece of paper.
 
This is not really a proof. But it might convince you.

Let
[tex]x=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...,[/tex]
then
[tex]2x=1+\frac{1}{2}+\frac{1}{}+\frac{1}{8}+\frac{1}{16}+...,[/tex]
so
[tex]2x-1=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...[/tex]

But the left-hand side is equal to x!

So [itex]2x-1=x[/itex]. Or equivalently [itex]x=1[/itex].
 
Curious3141 said:
From the way this post is worded, I'm assuming you don't want to hear a lot of technical jargon. So I'll keep it simple.

Just in case you weren't aware of the nomenclature (because I'll make reference to that):

Sequence = ordering of terms.

Series = Sum of terms in a sequence.

This is called a geometric series. This is the sum of a sequence of terms where the next term is derived by multiplying the current term by a fixed number (called the common ratio). So by knowing the first term and the common ratio, you can get all the terms of the sequence, and by then summing them, find the sum of the series.

Here the first term is 1/2 and the common ratio is also 1/2. Meaning the terms progressively get smaller. They will never completely vanish (become zero), but they will become arbitrary close as you take more and more terms.

This is called an infinite series because you "let" the series grow to an "infinite" length i.e. you don't stipulate an end to it. Practically, you will never be able to write down all the terms, but the continuation (...) at the end means that you're theoretically considering an infinite number of terms. As the sequence goes on, the terms get closer to zero, and this is important for something known as convergence, which means that the sum actually exists and is finite. If the series doesn't converge, it just "blows up" to arbitrarily larger numbers (this is called diverging to infinity).

In this case, because of the common ratio of 1/2, the terms get smaller and smaller and the series converges. We can work out the sum of the infinite geometric series as jgens stated (here it's [itex]\frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1[/itex] as your friend said). This essentially means that if you take more and more terms and sum them up (i.e. do the "partial sums" for a greater and greater number of successive terms), you will get arbitrarily close to 1, but never quite hit it. Here the partial sums will always be strictly less than one because the terms are all positive.

Hope this has made things very clear to you.
Great explanation~! I accually understood what you were saying :P Thank you very much, and to the poster that posted all of those #'s and stuff, I honestly had no idea what they meant, but I do appreciate the help, maybe explain all the mathematical stuff in a little detail? I know I have seen Ʃ before because I am currently self studying calculus and I believe that is called a piece wise function? yes? like it gives you 2 different things for a function...but not sure if that is the case here.
 
interesting trick micromass for the series.
 
This square of area 1

2q99937.jpg


should convince you that

1/2 + 1/4 + 1/8 + 1/16 + ... = 1

as we loosely say, and clarify what it means, that as we add more and more terms we get closer and closer to 1. That however many terms we include in the series we shall never get exactly to 1, but we shall get as close as we wish to 1 by taking enough terms. Therefore we can use this and similar seemingly ineffable statements for perfectly practical uses.
 
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jgens said:
Suppose [itex]0 \leq r < 1[/itex] and fix [itex]k \in \mathbb{N}[/itex]. Then we have the following:
[itex](1-r)\sum_{n=0}^k r^n = \sum_{n=0}^k r^n - r\sum_{n=0}^k r^n = 1-r^{k+1}[/itex]

I got really excited when I saw your post, Jgens. I have wanted to know how the formula for evaluating infinity sums was derived for a while. I don't understand the first part though.

How are you getting [itex]\sum_{n=0}^k r^n[/itex] to equal 1?
 
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