<s>[INST] $\epsilon-N$ proof for the limit of $a_n=n\left(\sqrt{1+\frac{1}{n}}-1\right)$ We are told the limit is 1/2, and I know the definition for $\epsilon-N$ convergence, and generally what I am looking for, but just cannot find the right algebra as scratchwork to find how to choose $N$ at the actual top of the proof. So far, what seems to be most useful is this
\begin{align}
\left| n \left(\sqrt{1+\frac{1}{n}}-1\right)-\frac{1}{2}\right|&=\left| \frac{n \left(\sqrt{1+\frac{1}{n}}-1\right) \left(\sqrt{1+\frac{1}{n}}+1\right)}{\sqrt{1+\frac{1}{n}}+1}-\frac{1}{2}\right|\\
&=\left| \frac{1}{\sqrt{1+\frac{1}{n}}+1}-\frac{1}{2}\right|
\end{align}
I have been chasing myself in circles with the rest of the algebra here to get to something that can obviously translate to a choice of $N$.
 [/INST] We have that $\frac{1}{\sqrt{1+\frac{1}{n}}+1}-\frac{1}{2} <0$ then we can consider, for $0<\varepsilon <\frac12$
$$\frac12-\frac{1}{\sqrt{1+\frac{1}{n}}+1} <\varepsilon \iff \frac{1}{\sqrt{1+\frac{1}{n}}+1} >\frac{1-2\varepsilon}{2}$$
$$\iff \sqrt{1+\frac{1}{n}}+1<\frac{2}{1-2\varepsilon} \iff \sqrt{1+\frac{1}{n}} <\frac{2}{1-2\varepsilon}-1$$
$$\iff 1+\frac{1}{n}<\left(\frac{1+2\varepsilon}{1-2\varepsilon}\right)^2 \iff \frac1n<\left(\frac{1+2\varepsilon}{1-2\varepsilon}\right)^2-1$$
$$\iff n>\frac1{\left(\frac{1+2\varepsilon}{1-2\varepsilon}\right)^2-1}$$
</s>