Why Does (r^n) Converge to 0?

[Lily@pie]

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

Homework Statement

I don't understand why does the sequence (rⁿ) converges to 0 as n -> infinity when -1<|r|<1

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.

 

[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.

 

[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.

 

[Lily@pie]

Ok. Thanks so much