ARQMATH LAB
Math Stack Exchange

Is this $\left|\left(\frac{a}{b}\right)^n-\left(\frac{a}{b}\right)^{n-1}\right|$ bounded?

3
$\begingroup$

Let $0.5 and let $b=1-a$. Let $n\in \mathbb{N}$.

$\left|\left(\frac{a}{b}\right)^n-\left(\frac{a}{b}\right)^{n-1}\right|\le C$.

Is $\left|\left(\frac{a}{b}\right)^n-\left(\frac{a}{b}\right)^{n-1}\right|\le C$ bounded by a constant $C$ for all $n$? If so, how would I show it/explain it?

analysis
asked 2012-05-27
user id:31569
451
11gold badges 55silver badges 1414bronze badges

2 Answers 2

4

From $$\frac 1 2 < a < 1$$ we get

$$0< b < \frac 1 2 $$

from where $$b1$$

Saying

$${\left( {\frac{a}{b}} \right)^n}\left| {1 - \frac{b}{a}} \right|$$

is bounded is the same as saying $${a_n} = {\left( {\frac{a}{b}} \right)^n}$$

is bounded.

Suppose

$${\left( {\frac{a}{b}} \right)^n} < R$$

for all $n$.

Since $a/b>1$, take $\log _{a/b}$. The inequality stays the same and

$$n < {\log _{a/b}}R$$

for all $n$. But this means $\Bbb N$ is bounded from above which is impossible.

asked 2012-05-27
user id:23350
98k
1010gold badges 151151silver badges 294294bronze badges
  • 0
    Thanks Peter, most appreciated! – 2012-05-27
  • 1
    Any time, Steven. =) – 2012-05-27
6

Well, you could try an example. Say $a = 0.8$ and $b = 0.2$. Then $a/b = 4$, so you are asking whether $4^n - 4^{n-1} = 4^{n-1} \cdot 3$ is bounded for all $n$. Is it?

asked 2012-05-27
user id:9019
2k
11gold badges 1111silver badges 1414bronze badges

Related Posts

No Related Post Found