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#### OhMyMarkov

##### Member

- Mar 5, 2012

- 83

Hello everyone!

I want to prove that $\lim \sqrt(n) \alpha ^n \rightarrow 0$ whenever $0 <\alpha < 1$. I got the following proof:

(1) Write $\alpha$ as $\alpha = 1/x$ where $x > 1$.

(2) $\sqrt{n} \alpha ^n = \displaystyle \frac{\sqrt{n}}{(1+x)^n}\leq\frac{\sqrt{n}}{1+nx}=\frac{1}{\frac{1}{\sqrt{n}}+x\sqrt{n}}\leq\frac{1}{x\sqrt{n}}\rightarrow 0$ as $n\rightarrow \infty$.

Is the proof I provided correct?

Thanks!

I want to prove that $\lim \sqrt(n) \alpha ^n \rightarrow 0$ whenever $0 <\alpha < 1$. I got the following proof:

(1) Write $\alpha$ as $\alpha = 1/x$ where $x > 1$.

(2) $\sqrt{n} \alpha ^n = \displaystyle \frac{\sqrt{n}}{(1+x)^n}\leq\frac{\sqrt{n}}{1+nx}=\frac{1}{\frac{1}{\sqrt{n}}+x\sqrt{n}}\leq\frac{1}{x\sqrt{n}}\rightarrow 0$ as $n\rightarrow \infty$.

Is the proof I provided correct?

Thanks!

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