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March 5, 2015

March 5, 2015

Posted by **mathstudent** on Tuesday, February 10, 2009 at 3:20pm.

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?

- math -
**bobpursley**, Tuesday, February 10, 2009 at 3:45pmSo 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) ?

- math -
**Damon**, Tuesday, February 10, 2009 at 3:48pmIf 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

- math -
**Damon**, Tuesday, February 10, 2009 at 3:49pmBy the way, if x is <1, then that x^n at the end will go to zero so it will all work.

- math -
**mathstudent**, Tuesday, February 10, 2009 at 4:36pmThanks 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.

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