Real Null Sequence/Examples/n^alpha over y^n

Example of Real Null Sequence

Let $\alpha \in \R$ be a (strictly) positive real number.

Let $y \in \R$ be a real number such that $\size y > 1$.

Let $\sequence {a_n}_{n \mathop \ge 1}$ be the real sequence defined as:

$\forall n \in \Z_{>0}: a_n = \dfrac {n^\alpha} {y^n}$

Then $\sequence {a_n}$ is a null sequence:

$\ds \lim_{n \mathop \to \infty} \dfrac {n^\alpha} {y^n} = 0$

Proof

Let $y = 1 + x$.

Then:

\(\ds \paren {1 + x}^n\)

\(=\)

\(\ds 1 + n x + \dfrac {n \paren {n - 1} } {2!} x^2 + \dfrac {n \paren {n - 1} \paren {n - 2} } {3!} x^3 + \cdots\)

General Binomial Theorem

\(\ds \)

\(>\)

\(\ds \dfrac {n^n x^n} {n!}\)

selecting the $n$th term

\(\ds \leadsto \ \ \)

\(\ds \dfrac {n^\alpha} {y^n}\)

\(<\)

\(\ds \dfrac {n^\alpha n!} {n^n x^n}\)

\(\ds \)

\(<\)

\(\ds \dfrac {n!} {n^{n - \alpha} \paren {y - 1}^n}\)

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Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $1$: Review of some real analysis: Exercise $1.5: 9$