Reciprocal of Null Sequence

Theorem
Let $\sequence {x_n}$ be a sequence in $\R$.

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

Then:
 * $x_n \to 0$ as $n \to \infty$ $\size {\dfrac 1 {x_n} } \to \infty$ 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 $n \to \infty$:
 * $\exists N: \forall n > N: \size {x_n} < H^{-1}$

That is:
 * $\size {\dfrac 1 {x_n} } > H$.

So:
 * $\exists N: \forall n > N: \size {\dfrac 1 {x_n} } > H$

and thus:
 * $\sequence {\size {\dfrac 1 {x_n} } }$ diverges to $+\infty$.

Suppose $\size {\dfrac 1 {x_n} } \to \infty$ as $n \to \infty$.

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

Also known as
Some sources call this the reciprocal rule, but as that name is used throughout mathematical literature for several different concepts, its use is not recommended.