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|$

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 
$$
\sqrt{1+\frac{1}{n}}-1 < 4\epsilon \implies n > \frac{1}{(4\epsilon +1)^{2}-1}
$$
