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oib111 · Posted: Sun May 03, 2009 12:00 pm Post subject: Infinite Geometric Sequences

How does an infinite geometric sequence have a sum? If it's infinite how can it's sum be constant?

Jani · Posted: Sun May 03, 2009 2:38 pm Post subject:

The ratio must be <1 and it works, trust me :D Anyway, you should be asking this your maths prof, not here, because it's way easier to explain the stuffs IRL and he should know hella lot more of maths than the majority of ppl here.

If you really want to read, Wikipedia has the info and Google has more.

It's very hard to write math stuff on a forum...

oib111 · Posted: Sun May 03, 2009 3:07 pm Post subject:

I know the ration has to be less than one. But it still doesn't make sense to me.... I'll read about it on wikipedia though.

Jani · Posted: Mon May 04, 2009 4:05 am Post subject:

Slugsnack · Posted: Tue May 05, 2009 8:14 am Post subject:

the point is that it doesn't sum to something. it tends to a limit instead, and you can find this limit. it's like if you keep halfing a number it will tend towards zero. or maybe multiplying it by negative half then it converges towards zero.

oib111 · Posted: Tue May 05, 2009 8:19 am Post subject:

I read wikipedia and I got it now. It was saying if you have:

S = 1 + 1/3 + 2/9 + 4/27 + ...,

And you multiply the first equation by 1/3 (the common ration) and then subtract the two equations everything cancels out except for the first term, so you get:

S - 2/3s = 1

You use Algebra to solve and you get S = 3. Therefore the sum is 3.

Trucido · Moderator Reputation: 6 Joined: 08 Sep 2007 Posts: 2792

I cant remember which one it is, but it ends up converging (or is it diverging?)
The rule is (for when r < 1)
Sn=(a(1-r^n))/1-r

You pretty much just flip it around for the r > 1.

Monkeys · Posted: Tue May 05, 2009 9:11 am Post subject: