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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.

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  • $\begingroup$ Try to prove this for the case when all $a_n$ are positive. If $a_n$ converges to zeros, what happens to $\frac{1}{a_n}$ $\endgroup$ – Sean Nemetz Apr 17 at 22:45
  • $\begingroup$ 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). $\endgroup$ – Winther Apr 17 at 22:51
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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.

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It's enough to show that $$\frac{1}{|a_n|}$$ is unbounded, which is clear since the denominator goes to 0.

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