# Convergence of r^n

1. Sep 25, 2011

### Lily@pie

I know this is like very basic, but my brain just somehow couldn't accept it!!

1. The problem statement, all variables and given/known data
I don't understand why does the sequence (rn) converges to 0 as n -> infinity when -1<|r|<1

3. The attempt at a solution
i did quite a few ways to convince myself.
Firstly, we know that (1/r)n<(1/r) only if 0<r<1. So by squeeze theorm, (1/r)n converges to 0. Then it also holds for -1<r<0 cz |(1/r)n| = (1/r)n.

this way seems to be correct but it doesn't seems to be convincing enough.

2. Sep 25, 2011

### CompuChip

If you want to make it more tangible, you could try it with some numbers:

If r = 0.1, then you get { 0.1, 0.01, 0.001, 0.0001, ... } and you see this quickly tends to 0.
If r = 0.5, then again { 0.5, 0.25, 0.125, 0.0625, ...}
Even for r = 0.99, { 0.99, 0.9801, ... } goes much more slowly, but 0.99100 is already of the order of 5 x 10-5.

Clearly, the boundary case is r = 1, as { 1, 1, 1, .... } never converges to 0.

For the rigorous proof, refer to your own post :-) What you did there is correct.

3. Sep 25, 2011

### deluks917

|r| < 1 so r = 1/x where |x|>1. r^n = 1/(x^n). If |x| >1 then clearly x^n becomes infinitely large as n goes to infinity. So 1/(big number) goes to zero.

4. Sep 25, 2011

### Lily@pie

Ok. Thanks so much