$$\lim_{n\to \infty} a_n=0 \iff \lim_{n\to \infty} \frac{a_{n+1}}{a_n}<1 \text{ when } a_n>0 \text{ and } \frac{a_{n+1}}{a_n}>0.$$
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There is a very simple counter example to this. – Cameron Williams Oct 21 '16 at 22:29
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The forward implication is not true. – dxiv Oct 21 '16 at 22:30
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The reverse implication is true. In fact, not only does $a_n \to 0$, but $\sum a_n < + \infty$. – MathematicsStudent1122 Oct 21 '16 at 22:39
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The other is d'Alembert rule. – hamam_Abdallah Oct 21 '16 at 22:39
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The forward implication is wrong; take for example $a_n=\frac 1 n, \lim_{n\to \infty} \frac {a_{n+1}} {a_n}=1$. The other implication however is true (if the limit of $\frac {a_{n+1}} {a_n}$ exists) and I used to call it "ratio criterion"; the proof is not difficult. First, note that $a_{n+1}<a_n$, so the sequence (which is also positive $\forall n\in \mathbb N$) has a limit. Moreover, you can say that$\frac {a_{n+1}} {a_n} < M $ for n big enough, where M is a constant $<1$. So $a_{n_0+k+1}< M a_{n_0+k} < M^2 a_{n_0+k-1}<..., a_1$ you can write $0< a_{n_0+k+1} < M^k a_{n_0} $, so $0< \lim_{k\to \infty} a_{n_0+k+1} < \lim_{k\to \infty} M^k a_{n_0}$, but we have M<1, so $a_n\to0$.
Bargabbiati
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How about this one: $$ a_n = \frac 1 n. $$ Then we have $$ \lim_{n\to\infty} a_n = 0 \text{ and } \lim_{n\to\infty} \frac{a_{n+1}}{a_n} = 1. $$ So the proposed theorem is false.
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Take
$$a_n=\begin{cases}\cfrac1n,&n\;\text{is odd}\\{}\\\cfrac1{2^n},&n\;\text{ is even}\end{cases}$$
Then $\;a_n\to0\;$ , yet
$$\frac{a_{n+1}}{a_n}=\begin{cases}\cfrac{2^n}{n+1},&n\;\text{is even}\\{}\\\cfrac n{2^{n+1}},&n\;\text{is odd}\end{cases}\;\;\;\;\implies\lim_{n\to\infty}\frac{a_{n+1}}{a_n}\;\;\text{doesn't even exist}$$
DonAntonio
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