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|$

$$n\left(\sqrt{1+\frac1n}-1\right)<n\left(\sqrt{1+\frac1n+\frac1{4n^2}}-1\right)=\frac12$$
and
$$n\left(\sqrt{1+\frac1n}-1\right)=\frac1{\sqrt{1+\dfrac1n}+1}>\frac1{\sqrt{1+\dfrac1n+\dfrac1{4n^2}}+1}=\frac12-\frac1{4n+1}.$$
