<s>[INST] How to bound above $| \frac{n}{\sqrt{n^2+n} +n}-\frac{1}{2} |$ 
Find and Justify $\lim\limits_{n \rightarrow \infty} S_n$
  $$S_n = \sqrt{n^2+n} - n = $$

$$\lim\limits_{n \rightarrow \infty} \sqrt{n^2+n} - n
 = \lim\limits_{n \rightarrow \infty} \frac{n}{\sqrt{n^2+n} +n}
 = \lim\limits_{n \rightarrow \infty} \frac{1}{\sqrt{1+1/n} + 1} = 1/2$$
$S_n$ converges to $\frac{1}{2}$ if for every $\epsilon>0 $ there is an integer $N$ s. t. 
$$|S_n-\frac{1}{2}| <\epsilon,n\geq N$$
Considering 
$$|S_n-\frac{1}{2}| = \left| \frac{n}{\sqrt{n^2+n} +n}-\frac{1}{2} \right|= \left|\frac{n-\sqrt{n^2+n}}{2\sqrt{n^2+n}+2n} \right|
\leq   \left|\frac{n-\sqrt{n^2+n}}{2n+2n} \right|
$$
$$ \leq 
\left|\frac{n-\sqrt{n^2+n}}{4n} \right| \leq
\left|\frac{1 -\sqrt{1+1/n}}{4}\right| \leq 
\frac{\sqrt{1/n}}{4} \leq \epsilon$$
let $N$ be such that 
$$ \frac{\sqrt{1/N}}{4} \leq \epsilon => \frac{1}{16 \epsilon^2}<N$$
It follows  for any $\epsilon>0$, if $n \geq N > \frac{1}{16 \epsilon^2}$, then 
$$|S_n-\frac{1}{2}| <\epsilon$$
I am a bit uncertain if it is the right way to proceed particularly with the bound of $N$ in relation to $\epsilon$. Is my method to bound $N$ correct?
Much appreciated for your input. 
 [/INST] Your method of proof for me it is correct.
Now the bound of $N$ you need   is $$N=[\frac{1}{16 \epsilon^2}]+1$$
This $N$ is the least natural number with the property:
$$  \frac{\sqrt{1/n}}{4} < \epsilon, \forall n\geqslant N $$
As for answering the comment below,you proved that $N> \frac{1}{16\epsilon^2}$.
We use the double inequality $[x]\leqslant x \leqslant [x]+1$ where $[x]$ is the integer part of $x$.
Thus we have $[\frac{1}{16\epsilon^2}] \leqslant \frac{1}{16\epsilon^2} \leqslant [\frac{1}{16\epsilon^2}]+1$
You see now that the the least positive natural number greater than $\frac{1}{16\epsilon^2}$ is $N=[\frac{1}{16\epsilon^2}]+1$
Thus whenever $n \geqslant N$ we have that desired inequality for the limit.
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