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

[andrey21]

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

 

[Dick]

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.

 

[andrey21]

Im sorry could u explain that a little more please.

 

[Dick]

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.

 

[andrey21]

well f(x) is x^n.

so differentiting gives:

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

 

[Dick]

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?

 

[andrey21]

s infinity = a/1-r correct?

 

[Dick]

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?

 

[andrey21]

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

 

[Dick]

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?

 

[andrey21]

Rite so doing that I obtain:

1/(1-x)

differentiate:

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

correct??

 

[Dick]

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?

 

[andrey21]

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

 

[Dick]

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?

 

[andrey21]

Well the right hand series is equivalent to:

n+1 (x)^n correct?

 

[Dick]

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?

 

[andrey21]

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

 

[Dick]

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

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

 

[andrey21]

Erm I don't see how that relates to:

n.(1/2)^n

Am i missing something obvious?

 

[Dick]

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.

 

[andrey21]

Well that gives:

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

1/2 / 1/4

= 2 correct??

 

[Dick]

Well that gives:

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

1/2 / 1/4

= 2 correct??

Correct.

 

[andrey21]

SO that is what the series converges to??

 

[Dick]

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??

 

[andrey21]

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