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

3
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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
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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
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  • 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
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