Show $ \lim_{n\rightarrow \infty} 2^{-1/\sqrt{n}}=1$ I am tasked with proving the following limit:
$$ \lim_{n\rightarrow \infty} 2^{-1/\sqrt{n}}=1$$
using the definition of the limit. I think I have done so correctly. I was hoping to have someone confirm my proof. Here is my reasoning:
We want 
$$ \left| 2^{-1/\sqrt{n}} - 1 \right| < \epsilon $$
for $\epsilon >0$ given. Rearranging, we have
$$ \left| 2^{-1/\sqrt{n}} - 1 \right| = \left| \frac{1-2^{1/\sqrt{n}}}{2^{1/\sqrt{n}}} \right| \leq \frac{1-2^{1/\sqrt{n}}}{2^{1/\sqrt{n}}} < \epsilon $$
by the Triangle Inequality and since we are forcing this quantity less than $\epsilon$. Rearranging again, we obtain
$$ 1 < 2^{1/\sqrt{n}}\left(1+\epsilon\right) $$
$$ \Rightarrow \ln \frac{1}{1+\epsilon} < \frac{1}{\sqrt{n}} \ln{2} $$
$$ \Rightarrow n > \left(\frac{\ln{2}}{\ln{\frac{1}{1+\epsilon}}} \right)^2 $$
where the inequality sign flipped since $ln\left(\frac{1}{1+\epsilon}\right)$ will be negative for all $\epsilon >0$. The proof should be straightforward:
Proof Let $\epsilon >0 $ be given. Define $N\left(\epsilon\right)=\left(\frac{\ln{2}}{\ln{\frac{1}{1+\epsilon}}} \right)^2$. Then,
$$ n>N\left(\epsilon\right) \rightarrow \cdots \rightarrow \left| 2^{-1/\sqrt{n}}-1\right| < \epsilon $$ QED.
Does this logic seem correct?
Thanks!
 A: Hint:  
Show that $|1 - (1/2)^{1/\sqrt{n}}| \leqslant 2/\sqrt{n}$ using the Bernoulli inequality.
A: $2^{-\frac{1}{\sqrt{n}}}={\mathrm e}^{\ln(\,2^{-\frac{1}{\sqrt{n}}})}={\mathrm e}^{-\frac{1}{\sqrt{n}}\ln(2)}=1+{\mathcal O}(\frac{1}{\sqrt{n}})$
A: Your logic is correct (minus a few minor problems).
If $n>0$, $2^{-1/\sqrt{n}}$ is always less than 1, so $2^{-1/\sqrt{n}}-1$ is always negative. Assuming $0<\epsilon< 1$,
$$\begin{align*}
\left|2^{-1/\sqrt{n}}-1\right|&=-(2^{-1/\sqrt{n}}-1)\\
&=1-2^{-1/\sqrt{n}}\\
&\leq\epsilon\\
\end{align*}$$
And now we can do some rearranging.
\begin{align*}
2^{-1/\sqrt{n}}\geq1-\epsilon\\
2^{1/\sqrt{n}}\leq \frac{1}{1-\epsilon}\\
\frac{1}{\sqrt{n}}\ln 2 \leq -\ln( 1-\epsilon)\\
n\geq \left(\frac{\ln 2}{\ln (1-\epsilon)}\right)^2
\end{align*}
A: The minus sign is not essential, we can remove it by inversion. We can assume $n$ to be a perfect square (as $2^{1/\sqrt n}<2^{1/\lfloor\sqrt n\rfloor}$) and study
$$\lim_{n\to\infty}2^{1/n}.$$
Then, 
$$\frac{2^{1/n}-1}{2-1}=\frac{2^{1/n}-1}{(2^{1/n})^n-1}=\frac1{(2^{1/n})^{n-1}+(2^{1/n})^{n-2}+\cdots1}<\frac1n.$$
