Infinite geometric series?

[Baron]

For what values of x , x ? 1, will the following infinite geometric series have a finite sum?
(x+1) + (x+1)^2 + (x+1)^3 ...

The answer is -2 < x < 0

Looking at the answer it seems that all coefficients when substituted x in (x+1) + (x+1)^2 + (x+1)^3 ... have to be between -1 < x < 1
How come?

 

[mmm4444bot]

Hello Baron:

The rule for an infinite geometric series

a + ar + ar^2 + ar^3 + …

is that a finite sum occurs when |r| < 1

So, in your exercise, we need |x + 1| < 1 because a = 1 and r = x + 1.

Using the definition of absolute value, we solve the compound inequality

-1 < x + 1 < 1

The solution is -2 < x < 0.

You mentioned coefficients. Perhaps, you're thinking of an infinite polynomial (I mean, the infinite expansion of the series), so maybe you're seeing those coefficients multiplied by powers of x and subsequently conclude -1 < x < 1.

I'm thinking that there is an issue when x is between 0 and 1; yes, the powers of x^n are getting smaller as n increases, but the coefficients coming out of expanding (x + 1)^n get so big so fast that x^n cannot eat away enough of them to get the partial sums to decrease (or something like that). This issue does not happen, when x is between -2 and 0.

Anyway, in the given infinite geometric series, I see coefficients as the first term (a = 1) and base as x + 1.

(x + 1) + (x + 1)^2 + (x + 1)^3 + (x + 1)^4 + …

We can factor out (x + 1):

(x + 1)[1 + (x + 1) + (x + 1)^2 + (x + 1)^3 + …]

The factor in square brackets is:

a + a(x + 1) + a(x + 1)^2 + a(x + 1)^3 + …

Visualizing this helps me realize -1 < x + 1 < 1.

Cheers ~ Mark