$\epsilon-N$ 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 $N$.
 A: Observe that
\begin{align}
\frac{1}{2}-n\left(\sqrt{1+\frac{1}{n}}-1\right) & = \frac{1}{2\left(\sqrt{n+1}+\sqrt{n}\right)^2}\le\frac{1}{8n}
\end{align}
I left the details to you.
A: Since $\frac{1}{\sqrt{1+1/n}+1}$ is $< 1/2$ and increasing in $n$, it suffices to find some $N$ such that $\frac{1}{\sqrt{1+1/N}+1} > \frac{1}{2} - \epsilon$. You can rearrange this inequality to solve for $N$.
A: We have that $\frac{1}{\sqrt{1+\frac{1}{n}}+1}-\frac{1}{2} <0$ then we can consider, for $0<\varepsilon <\frac12$
$$\frac12-\frac{1}{\sqrt{1+\frac{1}{n}}+1} <\varepsilon \iff \frac{1}{\sqrt{1+\frac{1}{n}}+1} >\frac{1-2\varepsilon}{2}$$
$$\iff \sqrt{1+\frac{1}{n}}+1<\frac{2}{1-2\varepsilon} \iff \sqrt{1+\frac{1}{n}} <\frac{2}{1-2\varepsilon}-1$$
$$\iff 1+\frac{1}{n}<\left(\frac{1+2\varepsilon}{1-2\varepsilon}\right)^2 \iff \frac1n<\left(\frac{1+2\varepsilon}{1-2\varepsilon}\right)^2-1$$
$$\iff n>\frac1{\left(\frac{1+2\varepsilon}{1-2\varepsilon}\right)^2-1}$$
A: So you have $\sqrt {n^2+n}-n$. Set $y=n+\frac 12$ then this becomes $\sqrt {y^2-\frac 14}-y+\frac 12$.
Now $$\sqrt {y^2-\frac 14}-y=\left(\sqrt {y^2-\frac 14}-y\right)\left(\frac{\sqrt {y^2-\frac 14}+y}{\sqrt {y^2-\frac 14}+y}\right)=\frac {\left(-\frac 14\right)}{\sqrt {y^2-\frac 14}+y}$$
I am sure you can fill in the gaps. This has a different flavour from some of the other suggestions. The strategy is to get a denominator which grows demonstrably faster than the numerator - here the numerator is a constant, but it can be done (in other cases of limits) with polynomials of different degrees - in that case, for high enough $N$ - a crude estimate is good enough - the highest term dominates.
