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Like the D'Alembert test case for convergence of series I have seen something similar for sequences in my math homework which required proof for this:

If we have a sequence $a_{n}$ such that $\space\lim_{n\rightarrow \infty}|\frac{a_{n+1}}{a_{n}}|<1$ then $\space\lim_{n\rightarrow \infty}a_n=0$.

How can I prove something like this without using the concept of convergency for series ? ( i.e. just using the Cauchy definition of the limit of a sequence) I have been trying to show that $|a_{n}|$ is a decreasing sequence but I don't know how to go further.

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Suppose we already know that $q^n \to 0$ if $|q| < 1$. If this is unclear, leave a comment and I'll post a proof for this as well.

If $\lim \left\vert \frac{a_{n+1}}{a_n} \right\vert < 1$, then there exists some $0 < c < 1$ and some $n_1 \in \mathbb N$ such that $$\left\vert \frac{a_{n+1}}{a_n} \right\vert < 1 - c$$ for all $n \geq n_1$.

Now let $\varepsilon > 0$. Since $(1-c)^n \to 0$, there exists some $n_2$ such that $$(1-c)^n < \frac{\varepsilon}{|a_{n_1}|}$$ for all $n \geq n_2$. Hence, for all $n > n_1 + n_2$, we have $$|a_n| = \left\vert a_{n_1} \right\vert \cdot \left\vert\frac{a_{n_1+1}}{a_{n_1}} \right\vert \cdot \ldots \cdot \left\vert\frac{a_n}{a_{n-1}} \right\vert < |a_{n_1}| \cdot (1-c)^{n - n_1} < |a_{n_1}| \cdot \frac{\varepsilon}{|a_{n_1}|} = \varepsilon$$ because $n-n_1 > n_2$.

So, what have we shown? For any $\varepsilon > 0$ there exists $N \in \mathbb N$ such that $$|a_n| < \varepsilon$$ for all $n \geq N$. Hence, $|a_n| \to 0$ and it is not hard to see that this implies $a_n \to 0$ as well.

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  • $\begingroup$ Thanks, this is really helpful. $\endgroup$ – Sota Antonino Dec 3 '15 at 15:35

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