Expert: Socrates - 8/27/2009

Question
QUESTION: Here's my question.

Consider the family of functions:

Fn(x) = x + 4x^2 + 9x^3 + ... + n^2*x^n

x = (0, 1)
n = 1, 2...

When n gets larger, which of the following functions if Fn getting close to?

a) x(1+x) / (1-x)^2

b) x(1+x) / (1-x)^3

c) x^2 (1+x) / (1-x)^2

d) X^2 (1+x) / (1-x)^3

when i plot the graphs for the x=(0,1) interval using the Maple software, they all look the same to me for that interval. how then do i know which answer is the correct one, and why?

Thank you so much :)

ANSWER: (Fn(x))/x = 1 + 4x + 9x^2 + ....+ (n^2) (x^(n-1))

= (x + 2x^2 + 3x^3 +....+ nx^n)'

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

= ((x) (x + x^2 + x^3 + ....+ x^n)' )'

Since  x + x^2 + x^3 + ....+ x^n approaches x/(1-x) as n gets larger ,

(x + x^2 + x^3 + ....+ x^n)' approaches 1/(1-x)^2

Then (x) (x + x^2 + x^3 + ....+ x^n)' approaches x/(1-x)^2

and ((x) (x + x^2 + x^3 + ....+ x^n)' )' approaches (x/(1-x)^2)' = (1+x)/(1-x)^3

So (Fn(x))/x approaches (1+x)/(1-x)^3

and

Fn(x) approaches (x)(1+x)/(1-x)^3

So the correct answer is b)


---------- FOLLOW-UP ----------

QUESTION: Thanks i understand better now.. it's something like making use of Mclaurin's series rite?

but for "Since  x + x^2 + x^3 + ....+ x^n approaches x/(1-x) as n gets larger", is there some name for this or is it just mathematically derived/worked out...? if it's the latter, how to go about doing it?

infinite thanks :)

Answer
" x + x^2 + x^3 + ....+ x^n approaches x/(1-x)"

x + x^2 + x^3 + ....+ x^n + ..... is a geometric series , with sum x/(1-x) , you should have seen this before you studied Mclaurin series.

If not , google "geometric series" , this is the first step to really understanding infinite series