Geometric Series

This is an example that you may already know something about. We have a formula for the sum of a simple finite geometrical series:

1 + x + x² + x³ + x⁴ + ^... + xⁿ = $${\frac{x^{n+1}-1}{x-1}}$$

(if x$\ne$1). Look up the proof of this if you do not know it already.

What about the infinite geometrical series?

1 + x + x² + x³ + x⁴ + ^... + xⁿ + ^...

According to our definition the sum of this series is the limiting value of its partial sums, if they have a limiting value. We already have a formula for the n^th partial sum:

s_n = $${\frac{x^{n+1}-1}{x-1}}$$

The question is: what happens to this as n$\to$$\infty$? The answer depends very much on the value of x. The only bit of the formula that depends on n is the x^{n + 1} in the numerator.

If x > 1 then x^{n + 1} tends to infinity as n$\to$$\infty$, so the partial sums also tend to infinity. So the series is divergent and does not have a sum.

If x < - 1 then the situation is even worse in a sense. Not only are the partial sums getting bigger and bigger, they are also switching sign. Certainly divergent.

If -1 < x < 1 then x^{n + 1}$\to$ 0 as n$\to$$\infty$ because the successive powers of a number between -1 and 1 get closer and closer to zero. So in this case our formula for the partial sums does tend to a finite limiting value as n$\to$$\infty$. So the infinite geometrical series is convergent in this case and has the sum

1 + x + x² + x³ + ^... + xⁿ + ^... = $${\frac{1}{1-x}}$$

The remaining two possibilities are x = 1 and x = - 1. I leave it to you to see that the series does not converge in either of these cases.

The end result is that the infinite geometric series given above only converges (has a sum) if -1 < x < 1.