Could someone explain the algebra in this sequences problem? $$ \frac{(n^2+n)-n^2}{\sqrt{n^2+n}+n} = \frac{n}{n\left(\sqrt{1+\frac{1}{n}}+1\right)} $$
The advantage here is that the n's cancel, but I don't see why the argument of the sqrt() function ends up becoming $$ 1 + \frac{1}{n} $$

You have $\sqrt {n^2 + n}$ what you are doing is factoring out $n^2$ from this expression which will give you 
$$\sqrt {n^2 \left( 1 + \frac{1}{n} \right)} = \sqrt {n^2 + \frac{n^2}{n}} = \sqrt{n^2 + n}$$
Multiplying out you'll see clearly that you'll end up with the factored expression. 
Since it is under the square root, the $n^2$ term simply becomes $n (\sqrt {n^2} = n)$
So that is how you end up with 
$$\frac{n}{n \sqrt {1+\frac{1}{n}} +n} = \frac{n}{n \left( \sqrt {1+\frac{1}{n}} +1\right)}$$
In this step you factor out $n$ from $n\sqrt {1+\frac{1}{n}}$ and $n$.
