# What does ( 0.25)^(3n) converge towards? (EASY Q)

1. Sep 30, 2009

### polosportply

Really quick question here:

I want to find the sommation of f(x) = 0.253n for x being a Natural ( N) number going from 0 to infinity.

k1k2n converges towards what, as a general rule?

Where k1, k2 are constants and n= 0,1,2,3 ...

EDIT: NO, not what it converges to, but what the sommation is equal to... as in f(0)+f(1) +f(2) + ...

Last edited: Sep 30, 2009
2. Sep 30, 2009

### Dick

It's a geometric series isn't it? It's ((1/4)^3)^n. What do you know about the sum of geometric series?

3. Sep 30, 2009

### polosportply

Well, I'm looking for = ∑rn-1 for n going from 1 to infinity... But isn't there a formula for where n starts at 0?

I also know that you can play around sith Sn= (1-rn) / (1-r) , but I don't remember/know where this leads to.

Last edited: Sep 30, 2009
4. Sep 30, 2009

### Dick

I'll help you out. The sum for n=0 to infinity of r^n is 1/(1-r) for r<1. That's the n->infinity limit of your Sn.

5. Sep 30, 2009

### polosportply

All right, thanks a lot

Last edited: Sep 30, 2009