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Suppose I have a monotonically decreasing sequence $a_{n}$ such that $a_{n}$ is positive for all $n \in \mathbb{N}^{+}$ and that

$$\lim_{n\rightarrow \infty} \frac{a_{n+1}}{a_{n}} = 0.$$

It seems to me that this is sufficient to say that $\lim a_{n} = 0$. Is this true? I tried to write a proof of the above proposition, but I feel a little uneasy about my argument (due to the contradiction and the funny stuff happening at 0) and I just want to make sure it isn't flawed.

By the monotone convergence theorem, suppose for a contradiction that $$\lim_{n\rightarrow \infty} a_{n} \rightarrow \alpha, \quad \alpha \in \mathbb{R}^{+}\setminus 0$$

It follows that $\lim_{n\rightarrow \infty} a_{n+1} \rightarrow \alpha$ and

$$\lim_{n\rightarrow \infty} \frac{a_{n+1}}{a_{n}} = \frac{\alpha}{\alpha} = 1.$$

This contradicts $\alpha \in \mathbb{R^{+}}\setminus {0}$ hence $\alpha = 0$.

I feel like the above should be encompassed in a more general statement about sequences and series in a standard calculus text, but I couldn't find it.

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    $\begingroup$ Proof is just fine. Just be a little bit more descriptive about the facts that $(a_n)$ converges since it is monotone decreasing and bounded below by $0$ and that all the $a_n$'s are positive and hence $ \dfrac{a_{n + 1}}{a_n} \to \alpha $ can be claimed using the assumption. Otherwise it is perfectly fine and is probably the best/shortest proof of the result. $\endgroup$ – Ishfaaq Sep 17 '14 at 1:04
  • $\begingroup$ I think the proof is lacking a rather important feature: to make it work you must first prove the limit of the sequence exists at all. Since the sequence is monotone descending thus it is necessary and sufficient to prove the sequence is bounded below $\endgroup$ – Timbuc Sep 17 '14 at 3:12
  • $\begingroup$ @Timbuc isn't that assumed from the hypothesis that $a_{n}$ is positive for each term in the sequence? $\endgroup$ – JessicaK Sep 17 '14 at 3:14
  • $\begingroup$ Of course, @JessicaK : I missed that part. Perhaps it'd be nice though to remark it again when assuming $\;a_n\to\alpha\neq 0\;$ . $\endgroup$ – Timbuc Sep 17 '14 at 3:16
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How about this:

Since

$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n} = 0$$

There exists some N such that for all $n>N$

$$ \left|\frac{a_{n+1}}{a_n}\right| < \frac{1}{2} $$

This means that

$$\left|a_{N+k}\right|<\frac{a_N}{2^k}$$

Which clearly converges to 0.

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One way to prove it is using series: since the infinite series

$$\sum_{n=1}^\infty a_n$$

is positive, we can apply the quotient test (D'Alembert's) to check convergence:

$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}$$

and since it is given the above limit exists and is less than one we get the infinite series converges, from which follows that $\;a_n\xrightarrow[n\to\infty]{}0\;$

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