Sum of a infinite series

1. Oct 16, 2012

phospho

$S = \frac{1}{2} + \frac{1}{4} + ... + (\frac{1}{2^n})$

I noticed that this is a sum of a infinite series with the common ratio being 1/2, so using $\frac{1}{1-1/2}$ I get S = 2, however with this question there is a hint saying multiply S by 2, which I did not use so I'm worrying if I done something wrong. Why is the hint there?

2. Oct 16, 2012

SammyS

Staff Emeritus
Re: Series

Assuming that n is some positive integer, then S is not an infinite series.

You can get the answer by looking at S as the difference of two infinite series with the common ratio being 1/2, one starting at 1/2, the other at 1/2n+1.

However, using the hint does a nice job of getting the result.

3. Oct 16, 2012

phospho

Re: Series

I don't get how you spotted that, nor how that works or why it's not a infinite series. The common ratio is a 1/2 and it seems to converge to something (2)?

4. Oct 16, 2012

Ray Vickson

Re: Series

If you wer to take n = 10 then you would have
$$\frac{1}{2} + \frac{1}{2^2} + \cdots + \frac{1}{2^{10}},$$ which is certainly not an infinite series. In fact, for any finite value of n you have a finite series.

RGV

5. Oct 16, 2012

phospho

Re: Series

OK

I'm still not being able to proceed

6. Oct 16, 2012

Ray Vickson

Re: Series

If the series WAS infinite, what would be its sum? (It is not 2, which is what you mistakenly stated in your original message.) What is the difference (literally) between the finite series and the infinite series?

RGV

7. Oct 16, 2012

rcgldr

Re: Series

$$2\ S = 1 + \frac{1}{2} + \frac{1}{4} + \ ... \ + \frac{1}{2^{n-1}}$$
$$1\ S = \ \ \ \ \ \ \ \frac{1}{2} + \frac{1}{4} + \ ... \ + \frac{1}{2^{n-1}} + \frac{1}{2^n}$$

What do you get if you subtract 1 S from 2 S?

8. Oct 16, 2012

Ray Vickson

Re: Series

Or, if we let S be the infinite series, we can write
$$S = \frac{1}{2} + \cdots + \frac{1}{2^n} = \frac{1}{2} + \cdots + \frac{1}{2^n} + \frac{1}{2^{n+1}} + \cdots - \left[ \frac{1}{2^{n+1}} + \cdots \right] = S_{\infty} -\frac{1}{2^n} S_{\infty} .$$

RGV

9. Oct 17, 2012

phospho

Re: Series

I've looked at your reply for 30 minutes and I don't know :\ I thought it was a infinite series, so I don't know how you are saying it can finite, and I don't know why it is not 2.

$$\frac{1}{2^n}$$

10. Oct 17, 2012

SammyS

Staff Emeritus
Re: Series

How did you cancel out the 1 that's in 2S ?

By the way:

$\displaystyle \sum_{k=1}^{\infty}\frac{1}{2^k}$ is an infinite series.

If n is a positive integer, then $\displaystyle \sum_{k=1}^{n}\frac{1}{2^k}$ is not an infinite series, it has a finite number, n, of terms.

11. Oct 17, 2012

rcgldr

Re: Series

Assume it's a finite series then consider what happens when the series become infinite. You missed the first term in 2S, which is the 1.

2S - 1S = 1 + (1/2 - 1/2) + (1/4 - 1/4) + ... + (1/2n-1 - 1/2n-1) - 1/2n = 1 + 0 + 0 + ... + 0 - 1/2n = 1 - 1/2n

So what is the limit as n → ∞ ?

12. Oct 17, 2012

phospho

Re: Series

Would it be 1?

$S_∞ = \frac{\frac{1}{2}}{1 - \frac{1}{2}}$ which is 1?

So if the sum to infinity is 1, and $S = 1 - \frac{1}{2^n}$ then what is the answer?

13. Oct 17, 2012

lendav_rott

Re: Series

This is your final answer. The sum gets ever closer to being 1, but it never will equal 1 because the elements in your sequence get smaller and smaller in value. You can try this out on a calculator if you're like me who always doubts everything :D

Or to test yourself wether you have understood the concept, you can take a random formula for a sequence, say
3/5^n , what does the sum of this sequence get closer to if n -> infinity?

Last edited: Oct 17, 2012
14. Oct 24, 2012

phospho

Re: Series

0.75

thanks to everyone who helped

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