Question

State whether this infinite series converges or diverges?

1+(t) + (t^3)+......

t=[(5x+6)/(3x-2)]

My thoughts on the question:

The sum of 'n' terms in a geometric progression is a[r^n - 1]/(r-1)--(let's calk this 1), where r>1 and when r<1, it is [a(1-(r^n)/(1-r) ---(let's call this 2)

Here a=(t^0) , r=t .But how do we decide which formula should be used from either 1 or 2,and apply the n-->(infinity) ,to find the above series converges or diverges?

Answer 1

We can see that t>1 for x>(-4) and t<1 for x<(-4)

Answer 2

If t = (5x+6)/(3x-2) then

|t| < 1
if -4 < x < -1/2

Answer 3

So we have to apply the two different formulae for the t>1 and t<1?

Answer 4

nope. It's just that the series converges iff |t| < 1

If you look at the formula, it's the same whether t<1 or t>1. The signs of the numerator and denominator both change, but the result is the same.

Answer 5

I missed that! Thank you!