I read from my textbook:

If S is the infinite series 1 + x + x^2 + x^3 + ...
Then Sx = x + x^2 + x^3 + x^4 + ... = S - 1
So, S = 1/(1-x)

I follow what that logic, but it still doesn't make sense.
The way I see it, if you plug any real number > 1 into x, S will be infinity which does not equal 1/(1-x) at all...

For example, if I plug the constant 10 in for x,
The infinite series "1 + x + x^2 + x^3 + ..." will be infinity
yet 1/(1-x) will equal -1/9.

Can someone explain this?

So if x>=1, both series do not converge, but diverge, and head for greater sums. But if x<1, they converge.

Now, if x<1, does S=1/(1-x) ?

If S is the infinite series 1 + x + x^2 + x^3 + ...

Then Sx = x + x^2 + x^3 + x^4 + ... = S - 1
well, say you terminate at 4 terms as you did:
xS = x + x^2 + x^3 + x^4
S =1+x + x^2 + x^3
then
xS-S = x^4 -1
and
S (x-1) = (x^4-1)
S = (x^4-1) / (x-1)
which is what you had except I have that x^4
No matter how many terms you take, xS will always have a sum bigger than S by that x^n at the end so it is not just the -1

By the way, if x is <1, then that x^n at the end will go to zero so it will all work.

Thanks guys. I wrote a simple computer program that verifies that S = 1/(1-x) holds when 0 < x < 1 (hence the series converges), but not when x > 1 (and the series diverges). That wasn't clear in the textbook. Thanks for the help.

The equation you mentioned, S = 1/(1-x), is called the geometric series formula, which is commonly used to find the sum of an infinite geometric series when the common ratio, x, lies between -1 and 1.

It is important to note that the formula only holds true when the absolute value of x is less than 1, so your observation is correct. When you plug in x = 10, which is greater than 1, the series no longer converges, and the formula does not apply. In this case, the sum of the infinite series is indeed infinity, as you correctly noticed.

Therefore, the formula S = 1/(1-x) only applies when the absolute value of x is less than 1. For example, if you plug in x = 1/2, the series becomes 1 + (1/2) + (1/2)^2 + (1/2)^3 + ..., and you can use the formula to find that the sum, S, is equal to 2.

In summary, to use the formula S = 1/(1-x) for finding the sum of an infinite geometric series, make sure that the common ratio, x, satisfies the condition |x| < 1. Otherwise, the series will not converge, and the formula will not apply.