Reciprocal of Null Sequence

Theorem
Let $$\left \langle {x_n} \right \rangle$$ be a sequence in $\mathbb{R}$.

Let $$\forall n \in \mathbb{N}: x_n > 0$$.

Then $$x_n \to 0$$ as $$n \to \infty$$ iff $$\frac 1 {x_n} \to \infty$$ as $$n \to \infty$$.

Proof

 * Suppose $$x_n \to 0$$ as $$n \to \infty$$.

Let $$H > 0$$.

So $$H^{-1} > 0$$.

Since $$x_n \to 0$$ as $$n \to \infty$$, $$\exists N: \forall n > N: x_n = \left|{x_n - 0}\right| < H^{-1}$$.

That is, $$\frac 1 {x_n} > H$$.

So $$\exists N: \forall n > N: \frac 1 {x_n} > H$$ and thus $$\left \langle {x_n} \right \rangle$$ diverges to infinity.


 * Suppose $$\frac 1 {x_n} \to \infty$$ as $$n \to \infty$$.

By reversing the argument above, we see that $$x_n \to 0$$ as $$n \to \infty$$.