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|$

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 $\frac{1}{4N}<\epsilon \rightarrow N>\frac{1}{4\epsilon}$
