Sequence of Powers of Reciprocals is Null Sequence

Theorem
Let $r \in \Q_{>0}$ be a strictly positive rational number.

Let $\sequence {x_n}$ be the sequence in $\R$ defined as:
 * $x_n = \dfrac 1 {n^r}$

Then $\sequence {x_n}$ is a null sequence.

Real Index
If $r \in \R_{>0}$ is a strictly positive real number, the same result applies.

However, the result is specifically stated for a rational index, as this definition is used in the course of derivation of the existence of a power to a real index.

Proof
Let $\epsilon > 0$.

We need to show that $\exists N \in \N: n > N \implies \left|{\dfrac 1 {n^r}}\right| < \epsilon$.

That is, that $n^r > 1 / \epsilon$.

Let us choose $N = \ceiling {\paren {1 / \epsilon}^{1/r}}$.

By Reciprocal of Strictly Positive Real Number is Strictly Positive and power of positive real number is positive, it follows that $\left({1/\epsilon}\right)^{1/r} \gt 0$.

Then by Positive Power Function on Non-negative Reals is Strictly Increasing:
 * $\forall n > N: n^r > N^r \ge 1 / \epsilon$.

Also see

 * Limit at Infinity of x^n