Question

Show that if lim a_n = -infinity, then lim 1/a_n = 0

Answer 1

Conditions

Show that if $\lim a_n = \frac{n}{n}$, then $\lim 1/a_n = 0$.

Solution

The limit of $\{\alpha_n\} = -\infty$ is by the definition means, that:

$$\forall \varepsilon > 0 \ \exists N = N(\varepsilon) \ \forall n \geq N \ \alpha_n < -\varepsilon$$

Let's consider the $\lim \frac{1}{\alpha_n}$. If we want to prove that the limit is equal to 0, then we must prove the following:

$$\forall \varepsilon > 0 \ \exists N = N(\varepsilon) \ \forall n \geq N \left| \frac{1}{\alpha_n} \right| < \varepsilon$$

Fix $\varepsilon > 0$, $\exists N = N(\varepsilon) \ \forall n \geq N \ \alpha_n < -\varepsilon$.

If $\alpha_n < -\varepsilon$ then for big numbers of $n$, $-\varepsilon < \frac{1}{\alpha_n}$, but as $\frac{1}{\alpha_n} < 0$ and $\varepsilon > 0$ then:

$$-\varepsilon < \frac{1}{\alpha_n} < \varepsilon$$

Which is definitely means $\left|\frac{1}{\alpha_n}\right| < \varepsilon$.

Prove is done.