What is the Limit of a Sequence with a Common Ratio of 1/2?

[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?

 

[SammyS]

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?

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.

 

[phospho]

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.

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)?

 

[Ray Vickson]

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)?

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

 

[phospho]

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

OK

I'm still not being able to proceed

 

[Ray Vickson]

OK

I'm still not being able to proceed

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

 

[rcgldr]

there is a hint saying multiply S by 2

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?

 

[Ray Vickson]

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?

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

RGV

 

[phospho]

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

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.

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?

\frac{1}{2^n}

 

[SammyS]

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?

...

\frac{1}{2^n}

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.

 

[rcgldr]

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?

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.

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) + ... + (1/2^n-1 - 1/2^n-1) - 1/2ⁿ = 1 + 0 + 0 + ... + 0 - 1/2ⁿ = 1 - 1/2ⁿ

So what is the limit as n → ∞ ?

 

[phospho]

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) + ... + (1/2^n-1 - 1/2^n-1) - 1/2ⁿ = 1 + 0 + 0 + ... + 0 - 1/2ⁿ = 1 - 1/2ⁿ

So what is the limit as n → ∞ ?

Would it be 1?

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

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?

 

[lendav_rott]

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?

 

[phospho]

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?

0.75

thanks to everyone who helped