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#### Alexmahone

##### Active member

- Jan 26, 2012

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Assume $\frac{a_{n+1}}{a_n}\to L$, where $L<1$ and $a_n>0$. Prove that

(a) $\{a_n\}$ is decreasing for $n\gg 1$;

I've done this part.

(b) $a_n\to 0$ (Give two proofs: an indirect one using (a), and a direct one.)

$\lim\frac{a_{n+1}}{a_n}=\frac{l}{l}=1$, a contradiction. So, $a_n\to 0$.

How do I go about the direct proof (one that doesn't use "proof by contradiction")?

(a) $\{a_n\}$ is decreasing for $n\gg 1$;

I've done this part.

(b) $a_n\to 0$ (Give two proofs: an indirect one using (a), and a direct one.)

__Indirect proof__: Assume for the sake of argument that $a_n$ does not tend to 0. Since the sequence is decreasing and bounded below by 0, it must converge to a positive value (call it $l$).$\lim\frac{a_{n+1}}{a_n}=\frac{l}{l}=1$, a contradiction. So, $a_n\to 0$.

How do I go about the direct proof (one that doesn't use "proof by contradiction")?

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