Real Null Sequence/Examples/n^alpha x^n/Lemma

Lemma for Real Null Sequence: $n^\alpha x^n$

Let $\alpha \in \Q$ be a (strictly) positive rational number.

Let $x \in \R$ be a real number such that $\size x < 1$.

There exists $N \in \N$ such that:

$\paren {1 + \dfrac 1 N}^{\alpha + 1} \, \size x \le 1$

Proof

Aiming for a contradiction, suppose:

$\forall n \in \N: \paren {1 + \dfrac 1 n}^{\alpha + 1} \, \size x > 1$

Then:

\(\ds \forall n \in \N: \, \)

\(\ds \dfrac 1 n\)

\(>\)

\(\ds \paren {\dfrac 1 {\size x} }^{1 / \paren {\alpha + 1} } - 1\)

\(\ds \)

\(>\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds \paren {\dfrac 1 {\size x} }^{1 / \paren {\alpha + 1} }\)

\(>\)

\(\ds 1\)

as $\dfrac 1 n > 0$ for all $n \in \N$

But this contradicts Sequence of Powers of Reciprocals is Null Sequence.

Hence by Proof by Contradiction:

$\exists N \in \N: \paren {1 + \dfrac 1 N}^{\alpha + 1} \, \size x \le 1$

$\blacksquare$

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 4$: Convergent Sequences: Exercise $\S 4.20 \ (5)$