Prove convergence of $n \cdot \left(\sqrt{1+ \frac{1}{n}} -1\right)$ I am working on some old analysis exams and i got stuck on this exercise : 

Using the epsilon definition show that $a_{n} = n \cdot \left(\sqrt{1+ \frac{1}{n}} -1\right)$ converges and determine its limit.

Knowing that the limit is 1/2, I know need to find an $ N \in \mathbb{N}$ so that : $ \forall \epsilon > 0 n > N \implies \left| a_n - \frac{1}{2} \right| < \epsilon $
Next step I simplify $a_n$ : $ a_n  = \frac{n \cdot \left(\sqrt{1+ \frac{1}{n}} -1\right) \cdot   \sqrt{1+ \frac{1}{n}} + 1 }  {\sqrt{1+ \frac{1}{n}} + 1} = \frac{1}{\sqrt{1+ \frac{1}{n}} + 1}$
And then I got stuck,what am I supposed to do with : $ \left|\frac{1}{\sqrt{1+ \frac{1}{n}} + 1} - \frac{1}{2} \right|$
 A: $$ a_n = \sqrt{n^2+n}-n = \frac{n}{\sqrt{n^2+n}+n} $$
is bounded between $$ \frac{n}{\left(n+\frac{1}{2}\right)+n}\quad\text{and}\quad\frac{n}{n+n} $$
so $\frac{1}{2}-a_n$ is bounded between
$$ 0\quad\text{and}\quad\frac{1}{2(4n+1)}.$$
A: You started well. You need to show that $ \left|\frac{1}{\sqrt{1+ \frac{1}{n}} + 1} - \frac{1}{2} \right|<\epsilon$ for $n>N$ where $N$ is some number.
$$ \left|\frac{1}{\sqrt{1+ \frac{1}{n}} + 1} - \frac{1}{2}\right|=\left|\frac{\sqrt n}{\sqrt{n+1} +\sqrt n} - \frac{1}{2} \right|=\left|\frac{2\sqrt n -\sqrt{n+1} -\sqrt n}{\sqrt{n+1} +\sqrt n}\right|=\left|\frac{\sqrt n -\sqrt{n+1}}{\sqrt{n+1} +\sqrt n}\right|=\left|\frac{1}{(\sqrt{n+1} +\sqrt n)^2}\right|<\left|\frac{1}{(\sqrt{n} +\sqrt n)^2}\right|=\frac{1}{4n}$$ 
Now we want to find such $N$ that $\frac{1}{4N}<\epsilon \rightarrow N>\frac{1}{4\epsilon}$
A: $$n\left(\sqrt{1+\frac1n}-1\right)<n\left(\sqrt{1+\frac1n+\frac1{4n^2}}-1\right)=\frac12$$
and
$$n\left(\sqrt{1+\frac1n}-1\right)=\frac1{\sqrt{1+\dfrac1n}+1}>\frac1{\sqrt{1+\dfrac1n+\dfrac1{4n^2}}+1}=\frac12-\frac1{4n+1}.$$
A: For a more accurate bound consider the following:
$$
\left|a_{n} - \frac{1}{2} \right| = \left|\frac{1}{\sqrt{1+\frac{1}{n}}+1} - \frac{1}{2} \right| = 
\left|\frac{2-(\sqrt{1+\frac{1}{n}}+1)}{2\left( \sqrt{1+\frac{1}{n}}+1 \right)} \right| \leq \left| \frac{2-(\sqrt{1+\frac{1}{n}}+1)}{4} \right| \implies \left| 2-(\sqrt{1+\frac{1}{n}}+1) \right| =  \left| 1-\sqrt{1+\frac{1}{n}} \right| = \left| \sqrt{1+\frac{1}{n}}-1 \right| = \sqrt{1+\frac{1}{n}}-1 < 4\epsilon 
$$
Then 
$$
\sqrt{1+\frac{1}{n}}-1 < 4\epsilon \implies n > \frac{1}{(4\epsilon +1)^{2}-1}
$$
