Is this $\left|\left(\frac{a}{b}\right)^n-\left(\frac{a}{b}\right)^{n-1}\right|$ bounded?
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analysis
2 Answers
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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.
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0Thanks Peter, most appreciated! – 2012-05-27
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1Any time, Steven. =) – 2012-05-27
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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?