What is the Convergence of the Geometric Series with Common Ratio 1/2?

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Homework Help Overview

The discussion revolves around the convergence of the series sum n*(1/2)^n, exploring its relationship to the geometric series and differentiation techniques to derive a formula for convergence.

Discussion Character

  • Exploratory, Mathematical reasoning, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the convergence of the series and the use of known geometric series results. There are attempts to differentiate the geometric series to derive a formula for the series in question. Questions arise about the manipulation of formulas and the application of the chain rule in differentiation.

Discussion Status

The discussion is active with participants engaging in mathematical reasoning and clarification of concepts. Some guidance has been provided regarding the differentiation of series and the application of derived formulas to the specific series in question. There is an ongoing exploration of how the derived results relate to the original series.

Contextual Notes

Participants are working within the constraints of homework rules, focusing on understanding the derivation and convergence without providing a final solution. There is an emphasis on confirming understanding rather than reaching a consensus.

andrey21
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Show what:
sum n.(1/2)^n
converges to
Now I know that sum (1/2)^n converges to 2 so can I use this some how
 
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Jamiey1988 said:
Show what:
sum n.(1/2)^n
converges to



Now I know that sum (1/2)^n converges to 2 so can I use this some how

Not really, but knowing what sum x^n is and taking a derivative with respect to x would help.
 


Im sorry could u explain that a little more please.
 


Jamiey1988 said:
Im sorry could u explain that a little more please.

1+x+x^2+x^3+...=f(x) for |x|<1. What is f(x)? Now differentiate both sides.
 


well f(x) is x^n.

so differentiting gives:

n.x^(n-1) = 1+ 2x + 3x^2 + 4x^3...
 


Jamiey1988 said:
well f(x) is x^n.

so differentiting gives:

n.x^(n-1) = 1+ 2x + 3x^2 + 4x^3...

f(x) isn't x^n! f(x) is the sum of the geometric series x^n. f(x)=1+x+x^2+x^3+x^4+... out to infinity. There's no 'n' in the answer. What formula do you use to sum an infinite geometric series?
 


s infinity = a/1-r correct?
 


Jamiey1988 said:
s infinity = a/1-r correct?

a/(1-r) is right if the series is a*r^n. What is it if the series is x^n?
 


Am I meant to manipulate the existing formula or is there a separate one, if so I havnt come across it.
 
  • #10


Jamiey1988 said:
Am I meant to manipulate the existing formula or is there a separate one, if so I havnt come across it.

Manipulate the existing formula. x^n is a*r^n when a=1 and r=x, isn't it?
 
  • #11


Rite so doing that I obtain:

1/(1-x)

differentiate:

1.(1-x)^-1
-1.(1-x)^-2

correct??
 
  • #12


Jamiey1988 said:
Rite so doing that I obtain:

1/(1-x)

differentiate:

1.(1-x)^-1
-1.(1-x)^-2

correct??

Close, but no. You are forgetting the chain rule part of the calculation. d/dx (u)^(-1)=(-1)*u^(-2)*du/dx. What's the du/dx part?
 
  • #13


Ah so would that make it (1-x)^-2??
 
  • #14


Jamiey1988 said:
Ah so would that make it (1-x)^-2??

Ok. So put the two things together. 1/(1-x)^2=1+2*x+3*x^2+4*x^3+... Now what's the difference between the series on the right side and the series you want to sum n*x^n?
 
  • #15


Well the right hand series is equivalent to:

n+1 (x)^n correct?
 
  • #16


Jamiey1988 said:
Well the right hand series is equivalent to:

n+1 (x)^n correct?

Right. Suppose you multiply both sides by x? Does that get you closer to what you want?
 
  • #17


So that would give x/(1-x)^2 = x + 2x^2 +3x^3...
 
  • #18


Jamiey1988 said:
So that would give x/(1-x)^2 = x + 2x^2 +3x^3...

Ok, so might that help you to solve the given problem?
 
  • #19


Erm I don't see how that relates to:

n.(1/2)^n

Am i missing something obvious?
 
  • #20


Jamiey1988 said:
Erm I don't see how that relates to:

n.(1/2)^n

Am i missing something obvious?

Yes, you are missing something REALLY obvious. Put x=1/2 in the formula you just derived.
 
  • #21


Well that gives:

1/2 /(1-1/2)^2

1/2 / 1/4

= 2 correct??
 
  • #22


Jamiey1988 said:
Well that gives:

1/2 /(1-1/2)^2

1/2 / 1/4

= 2 correct??

Correct.
 
  • #23


SO that is what the series converges to??
 
  • #24


Jamiey1988 said:
SO that is what the series converges to??

Why would you think not?? Isn't the series 1*(1/2)^1+2*(1/2)^2+3*(1/2)^3+ etc? Isn't that sum n*x^n for x=(1/2)? Didn't we show that is x/(1-x)^2? Where along the line did I lose you??
 
  • #25


No I was just confirming that's all. Thank u for all ur help :)
 

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