1. ## Convergence Proof

I need to use the Archimedean Principle to prove that

$\displaystyle 2^{-n} and \frac{1}{\sqrt{n}}$

both converge to zero.

Using this principle it tells me I can choose N $\displaystyle \epsilon$ N such that N > something $\displaystyle \varepsilon$ . This is where I get stuck because I dont know what to choose for something $\displaystyle \varepsilon$ in either of the cases. Maybe the fraction I could use $\displaystyle \frac{1}{\varepsilon}$ . Any suggestions??

2. Here are strong hints for all three.

The Archimedean Principle says: $\displaystyle \varepsilon > 0 \Rightarrow \left( {\exists N \in \mathbb{N}} \right)\left[ {\frac{1}{N} < \varepsilon } \right]$.
So that means $\displaystyle n > N \Rightarrow \,\frac{1}{n} \leqslant \frac{1}{N} < \varepsilon$.

$\displaystyle \begin{gathered} \varepsilon > 0 \Rightarrow \varepsilon ^2 > 0 \Rightarrow \left( {\exists N \in \mathbb{N}} \right)\left[ {\frac{1}{N} < \varepsilon ^2 \Rightarrow \frac{1}{{\sqrt N }} < \varepsilon } \right] \hfill \\ n > N \Rightarrow \,\frac{1}{{\sqrt n }} \leqslant \frac{1}{{\sqrt N }} < \varepsilon \hfill \\ \end{gathered}$

$\displaystyle 2^n \geqslant n \Rightarrow \frac{1}{{2^n }} < \frac{1}{n} < \varepsilon$.