<s>[INST] 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|$
 [/INST] $$ a_n = \sqrt{n^2+n}-n = \frac{n}{\sqrt{n^2+n}+n} $$
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
$$ 0\quad\text{and}\quad\frac{1}{2(4n+1)}.$$
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