Prove: If $|r| < 1$, then $\lim_{n\to \infty }r^n$ = 0.

Question

Prove: If $|r| < 1$, then $\lim_{n\to \infty} r^n = 0$.

I know I am supposed to use the fact that $\lim_{n\to \infty} r^n = 0$ iff $\lim_{n\to \infty} |r^n| = 0$. I also know that $|r^n|$ is monotone and that I have to prove it by plugging in a value like 1/2 to show it's decreasing but that is as far as I have gotten. I just do not know how to use those ideas. I am stuck on how to prove that it is monotone and how the proof should look in general. Any help would be much appreciated.

Does this answer your question? Limit of the geometric sequence ${r^n}$, with $|r| < 1$, is $0$? — Martin R, Sep 27 '20 at 05:50

score 0 · Answer 1 · edited Sep 27 '20 at 06:03

0

Welcome to MSE. this is an idea $$|r|<1\implies\\ 1>|r|>|r^2|>|r^3|>...\geq0$$ for the case of $r^n$ you can rewrite $$0<r<1 \implies 1>r>r^2>...\geq0\\and\ \\ \text{case 1 }\\-1<r<0\implies 0\geq...>r^{2n+3}>r^{2n+1}>...>r^3>r\\\text{case 2}\\ r^2>r^4>r^6>...>r^{2n}>r^{2n+2}>...\geq0 $$

edited Sep 27 '20 at 06:03

Calvin Khor

34,903

answered Sep 27 '20 at 05:55

Khosrotash

24,922

But a positive decreasing seqence does not necessarily converge to zero. – Martin R Sep 27 '20 at 05:56
That just shows it's decreasing. Not thatt it converges to $0$. – fleablood Sep 27 '20 at 05:59
This can easily be fixed, just writing $|r^{n+1}| = |r| \times |r^n|$, and taking the limit, you get $l=|r| l$ so $l=0$. – TheSilverDoe Sep 27 '20 at 14:00

Prove: If $|r| < 1$, then $\lim_{n\to \infty }r^n$ = 0.

1 Answers1