Convergence of Taylor Series of Function Analytic on Disk/Lemma/Proof 2
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Lemma
Let $y > 1$.
Then:
- $\ds \lim_{n \mathop \to \infty} \frac n {y^n} = 0$
Proof
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 $\ln y > 0$ |
$\blacksquare$