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

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

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→∞) Ax+1/Ax = 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 + Ar2 + Ar3 + Ar4...       [First equation]

If you multiply both sides by r, you get:

Sr =Ar + Ar2 + Ar3 + Ar4 + Ar5...

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.