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Convergence Proof (As Part of Geometric Series Sum)

  • Thread starter Mathmo
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  • #1
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Homework Statement


I am trying to prove the sum of a geometric series, but one of the steps involves deriving this result:

[tex]\lim_{n\to\infty}r^{n}=0[/tex]

so that you can simplify the sum of a geometric series, where I have got to this stage:

[tex]S_{\infty} = \frac{a(1-r^{\infty})}{1-r}[/tex]


Homework Equations


[tex]\lim_{n\to\infty}r^{n}=0[/tex]
s.

The Attempt at a Solution


I've managed to do the rest of the derivation and can continue past the above steps, by assuming that the limit does equal zero, but I am stuck on the proof. I've looked online and it seems you need calculus to prove this, but we've not been taught any.

I know the limit equals zero for r <|1|, as it makes sense intuitively, but how do I prove this or start to?

Thanks
 

Answers and Replies

  • #2
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Uuh, how can you talk about series without knowing calculus in the first place...

Anyway, you have to show this limit is 0. So, what does "limit" mean?? What is the definition??
 
  • #3
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The limit is what it tends to, I understand that and know it tends to zero by trying values. e.g. (1/2)^9999, but how do I prove this?
 
  • #4
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You prove that by applying the definition of limit. What is that??
 
  • #5
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[tex]\lim_{x\to a}f(x)=L[/tex] means that given any real number e>0, there exists another real number d, such that:

if 0<|x-a|<d then |f(x) - L|<e

But I'm not sure what to do next?
 
  • #6
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3,279
[tex]\lim_{x\to a}f(x)=L[/tex] means that given any real number e>0, there exists another real number d, such that:

if 0<|x-a|<d then |f(x) - L|<e

But I'm not sure what to do next?
That is the definition for the limit of a continuous function. Here you want to do another limit: you want to take the limit of a sequence. This is something completely different then what you wrote down...

The thing you need to prove is that for all e>0, there existss a positive integer N, such that for all n>N holds [itex]|r^n|<e[/itex].
 
  • #7
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Thanks for your help, but I must just be really stupid, as I'm still stuck on how to progress. :(
 
  • #8
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Let 0<r<1. We can write r=1/(1+p) for p>1. Now we can do

[tex](1+p)^n> np[/tex]

So

[tex]0<r^n<\frac{1}{n p}[/tex]

So in order to prove that the left hand side converges to zero, it suffices to prove that the right hand side converges to 0.
 
  • #9
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Thanks again for your help, but now I'm even more confused:

1) Do you mean p > 0?
2) What allows you to say that (1+p)^n >np
 
  • #10
HallsofIvy
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I think that we should point out that your basic statement, [itex]\lim_{n\to\infty} r^n= 0[/itex] is only true for some values of r. What are your conditions on r?
 
  • #11
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Thanks again for your help, but now I'm even more confused:

1) Do you mean p > 0?
I mean that p is strictly larger than zero.

2) What allows you to say that (1+p)^n >np
Prove this by induction.
 
  • #12
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I think that we should point out that your basic statement, [itex]\lim_{n\to\infty} r^n= 0[/itex] is only true for some values of r. What are your conditions on r?
-1<r<1
I think

I mean that p is strictly larger than zero.



Prove this by induction.
I will try to prove this by induction and come back.
 
  • #13
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Let 0<r<1. We can write r=1/(1+p) for p>1. Now we can do

[tex](1+p)^n> np[/tex]

So

[tex]0<r^n<\frac{1}{n p}[/tex]

So in order to prove that the left hand side converges to zero, it suffices to prove that the right hand side converges to 0.
Again thanks for your help, I can show the first part of induction, that for n=1:

[tex](1+p) > p[/tex]

which is true.

When I try and extend this to n+1 I get:

[tex](1+p)^{n+1} > (n+1)p[/tex]

[tex](1+p)^{n}(1+p) > np + p[/tex]

But I don't know how to prove this last step?
 
  • #14
22,097
3,279
Again thanks for your help, I can show the first part of induction, that for n=1:

[tex](1+p) > p[/tex]

which is true.

When I try and extend this to n+1 I get:

[tex](1+p)^{n+1} > (n+1)p[/tex]

[tex](1+p)^{n}(1+p) > np + p[/tex]

But I don't know how to prove this last step?
What is your induction hypothesis?? Multiply both sides of it by 1+p.
 

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