Why is the limit below equal to $0$ ? It is known that $a>1$? $$\lim_{n\to \infty} \frac{n}{a^n}$$
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Hint If $ a> 1$ and $n \geq 2$ then $$a^n =(1+(a-1))^n > \frac{n(n-1)}{2} (a-1)^2$$
N. S.
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Hint: you can use L'Hospital to solve the limit, I figured it out, but thank you! – Maier Tudor Nov 05 '15 at 18:28
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1@MaierTudor L'H works but in my opinion is an overkill. – N. S. Nov 05 '15 at 18:33
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Let $(x_{n})=\frac{n}{a^{n}}$ and consider $\frac{x_{n+1}}{x_{n}}=\frac{n+1}{an}=\frac{1}{a} +\frac{1}{an}$. Taking the limit as $n\rightarrow \infty$, the second term goes to zero so we are left with $\lim \frac{x_{n+1}}{x_{n}} =\frac{1}{a} >1$ as $a>1$, therefore, by the theorem sometimes referred to as the "Cauchy Ratio Test," $\lim \frac{n}{a^{n}}=0$.
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Notice, given $a>1$ $$\lim_{n\to \infty}\frac{n}{a^n}=\lim_{n\to \infty}\frac{\underbrace{n}_{\large \to \infty}}{\underbrace{a^n}_{\large \to \infty}}$$ Applying L'Hospital's rule, $$=\lim_{n\to \infty}\frac{\frac{d}{dn}(n)}{{\frac{d}{dn}(a^n)}}=\lim_{n\to \infty}\frac{1}{na^{n-1}}=\lim_{n\to \infty}\frac{1}{n}\cdot \lim_{n\to \infty}\frac{1}{a^{n-1}}=(0)(0)=\color{red}{0}$$
Harish Chandra Rajpoot
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3Incorrect derivative in the denominator. Mind you, L'Hospital feels like overkill here. – Jyrki Lahtonen Nov 05 '15 at 19:21