Convergence of Taylor Series of Function Analytic on Disk/Lemma

Lemma

Let $y > 1$.

Then:

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

Proof 1

Let $z$ be a real number.

We have:

$\ds \sum_{n \mathop = 0}^\infty n z^n$ has radius of convergence of $1$

by Complex Power Series/Examples/n.

This leads to that:

$\ds \sum_{n \mathop = 0}^\infty n \paren {\frac 1 y}^n$ converges

as $\ds \frac 1 y < 1$ as $y > 1$.

This leads to that:

the sequence $\ds \sequence {\frac n {y^n}}$ converges to the limit $0$

by Terms in Convergent Series Converge to Zero.

This means that:

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

by definition of limit.

This concludes the proof.

$\blacksquare$

Proof 2

Note that $\ln y > 0$ as $y > 1$.

\(\ds \lim_{n \mathop \to \infty} \frac n {y^n}\)

\(=\)

\(\ds \lim_{n \mathop \to \infty} \frac n {\paren {e^{\ln y} }^n}\)

\(\ds \)

\(=\)

\(\ds \lim_{n \mathop \to \infty} \frac n {e^{\paren {\ln y} n} }\)

\(\ds \)

\(=\)

\(\ds 0\)

as $\ds \lim_{x \mathop \to \infty} \frac x {e^{\paren {\ln y} x} } = 0$ by Limit at Infinity of Polynomial over Complex Exponential as $\ln y > 0$

$\blacksquare$