Let $0.5<a<1$ 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?


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

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

from where $$b<a \Rightarrow \frac{a}{b}>1$$


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

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

is bounded.


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

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.

  • $\begingroup$ Thanks Peter, most appreciated! $\endgroup$ – Steven May 27 '12 at 23:18
  • 1
    $\begingroup$ Any time, Steven. =) $\endgroup$ – Pedro Tamaroff May 27 '12 at 23:26

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?


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