Property of limits explanation

Question

Why when we have a sequence $\left\lbrace a_{n}\right\rbrace$ and we know that $$ \frac{a_{n+1}}{a_n} < 1 \qquad \text{for big enough }n \text{ then } \qquad \lim\limits_{n \to \infty} a_n = 0 $$ Do we have any theorem for this property?

Answer 1

We don't know that, and we can't, because it isn't true.

Take for example $a_n = \frac{n}{n-1}$, then

$$\frac{a_{n+1}}{a_n} = \frac{\frac{n+1}{n}}{\frac{n}{n-1}} = \frac{n^2-1}{n^2} < 1$$ however $$\lim_{n\to\infty} a_n = 1,$$

proving a counterexample.

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What we do know is that if there exists some $c<1$ such that $\frac{a_{n+1}}{a_n} < c$ for all large enough $n$, then $a_n\to 0$. We know this because, if the inequality is true for all $n>N$, and if we know that $k>N$, we have

$$a_{k} < c\cdot a_{k-1}<c^2\cdot a_{k-2}<\cdots <c^{k-N}a_N$$ and this clearly goes to $0$.

Answer 2

No, we do not have a theorem for that, because the statement is worng. Consider for example $a_n = 1 +\frac 1n$.