# Proving $\lim _{n\to\infty}\frac{1}{ a_n} \neq \alpha$ given $\lim _{n\to\infty}a_n = 0$ and…

Given: $$\displaystyle {\lim _{n\to\infty}}a_n = 0\\\alpha \in \mathbb{R}\\a_n \neq 0$$ I'm trying to show:
$$\exists \mathcal{E} > 0| \exists N \in \mathbb{N}|\forall n > N:$$ $$\left| \frac{1}{a_n} - \alpha\right| \geq \mathcal{E}$$

I took the following steps: $$\left| \frac{1}{a_n} - \alpha\right| = \left| \frac{1 - \alpha \cdot a_n}{a_n}\right|$$

From here, I split the proof into cases: $$\alpha = 0$$, $$\alpha >0$$, and $$\alpha < 0$$. I managed to prove it for $$\alpha = 0$$, where the absolute value is always larger than zero and therefore allows for the inequality for every $$n$$. But when trying to prove it for the other cases, say, $$\alpha >0$$:

$$\left| \frac{1 - \alpha \cdot a_n}{a_n}\right| = \left| \frac{1}{a_n}-\alpha\right|\geq \mathcal{E}$$ How do I proceed from here? I don't know how to get a simpler inequality beyond this point.

• Try to prove this for the case when all $a_n$ are positive. If $a_n$ converges to zeros, what happens to $\frac{1}{a_n}$ – Sean Nemetz Apr 17 '19 at 22:45
• It's much easier to start with the definition of $a_n \to 0$ and use this to show that the magnitude of $\frac{1}{a_n}$ is eventually "very large" (in a way you can quantify for the $\epsilon$ you get too choose in the definition). – Winther Apr 17 '19 at 22:51

For the case $$\alpha >0$$ use triangle inequality: it is enough to show that $$|\frac 1 {a_n}| >\epsilon +\alpha$$. (Because $$|\frac 1 {a_n}-\alpha|\geq |\frac 1 {a_n}| -\alpha$$). Hence we only need $$|a_n| <\frac 1 {\alpha +\epsilon}$$ which is true for $$n$$ sufficiently large since $$a_n \to 0$$. The case $$\alpha <0$$ can be handled similarly.
It's enough to show that $$\frac{1}{|a_n|}$$ is unbounded, which is clear since the denominator goes to 0.