**URGENT: CALC HELP** Involves Power series...

[mAdD INDIAN]

Ok after a weekend of calc studying (and 7 hours today), I have come to the conclusion that I am still gonna fail after looking at the past exams.

Anyway I have a question regarding power series:
How do you find the exact sum of SIGMA< k/3^k) as k--> infinity ?

In general how would you find the sum for the above type of problems. I know about the 1/1-r identity if r<1, but I can't seem to apply it in this case??

 

[Martin]

hold on, let me type in my answer...

 

[Martin]

I think I know what you're talking about. what you have to do is make that series look like a known power series, then find the sum.


Example:

Sigma (from 1 to infinity) = n/3^n

1. write it out as n*(1/3)^n
2. replace 1/3 with x and you get Sigma n*x^n

You should be able to find this sum, by letting S(x) = sigma n*x^n and performing operations to the right hand side to look like a standard power seres (x^n in this case)

when you finally get an expression for S(x) { x/(x-x)^2 in this case } you once again replace the x with a 1/3

 

[Spac3d]

I love power series!!

 

[mAdD INDIAN]

Originally posted by: MartyTheManiak
I think I know what you're talking about. what you have to do is make that series look like a known power series, then find the sum.


Example:

Sigma (from 1 to infinity) = n/3^n

1. write it out as n*(1/3)^n
2. replace 1/3 with x and you get Sigma n*x^n

You should be able to find this sum, by letting S(x) = sigma n*x^n and performing operations to the right hand side to look like a standard power seres (x^n in this case)

when you finally get an expression for S(x) { x/(x-x)^2 in this case } you once again replace the x with a 1/3

U mean use the formula SIGMA ( x^n ) = 1/(1-x)?

So you get that and multiply n to it, so its n/(1-x) which becomes n/(1-1/3) ??