in a geometric sequence, the generator is the number that one term is multiplied by to generate the next term.

## what happens to the tearms of sequence when the multiplier is less than 1, but greater than 0?

# 2 Answers

Well, for example, your common ratio (generator) could be 0.5, which is less than 1, but greater than 0. Let us look at what happens with different starting values.

**80**: 80, 40, 20, 10, 5, 2.5, 1.25, .625, .3125, ....

**200: **200, 100, 50, 25, 12.5, 6.25, 3.125, ...

**1024: **1024, 512, 256, 128, 64, 32, 16, 8, 4, 2, 1, .5, .25, .125, ...

We can pick any number of examples with different starting values and different common ratios between 0 and 1, but the idea is that in each case, the numbers would always get smaller (since we are multiplying by a number less than 1), but never change sign
from positive to negative (since we are multiplying by a positive number), so **
our terms would get closer and closer to zero, but never reach it.**

The geometric series you describe converges. This can be proven with the ratio test.

The ratio test states that if the ratio of succeeding terms is a constant that is less than zero, the series converges.

lim(_{x→∞}) A_{x+1}/A_{x} = r

If 0<r<1, then the series converges.

The exact value of a convergent, geometric series can often be found.

Let r be the ratio between two terms. For example, in the series 9, 3, 1, 1/3, 1/9, the ratio is 1/3.

For a starting number A, the sum of the sequence (S) is:

S = A + Ar + Ar^{2} + Ar^{3} + Ar^{4}... [First equation]

If you multiply both sides by r, you get:

Sr =Ar + Ar^{2} + Ar^{3} + Ar^{4} + Ar^{5}...

Now subtract this from the first equation:

S-Sr = A

All the other terms cancel out.

Solve for S:

S(1-r) = A

S = A / (1-r)

If your ratio is 1/2 and you start at 1, then the sum of the infinite series is S = 1 /(1-1/2) = 2

You can plug any starting number A and any ratio r into the equation to get the sum IF the series converges, which it will IF the ratio is between 0 and 1.