How is $\sqrt{n^2+2n}-\lfloor\sqrt{n^2+2n}\rfloor=\frac{2}{\sqrt{1+\frac{2}{n}}+1}$. I have trouble seeing how
$$\sqrt{n^2+2n}-\lfloor\sqrt{n^2+2n}\rfloor=\frac{2}{\sqrt{1+\frac{2}{n}}+1}.$$
I can't see where to start even.
 A: Assuming that $n$ is a non-negative integer, we know that 
$$ n^2 \leq n^2 + 2n < n^2 + 2n + 1 = (n+1)^2$$
Hence, this show that $\lfloor \sqrt{n^2 + 2n} \rfloor = n$, which simplifies the LHS to
$$ \sqrt{n^2+2n} - n.$$
Now let's work on simplifying the RHS, multiplying the numerator and denominator by $n$, we get
$$ \frac {2}{ \sqrt{1+ \frac {2}{n}} + 1} \times \frac {n}{n} = \frac {2n} {\sqrt{n^2 +2n} +n}$$
Let's also rationalize the denominator, by multiplying throughout by $\frac {\sqrt{n^2+2n} -n}{\sqrt{n^2+2n}-n}$ to obtain
$$\frac {2n} {\sqrt{n^2 +2n} +n} \times \frac {\sqrt{n^2+2n} -n}{\sqrt{n^2+2n}-n} = \frac {(2n) ( \sqrt{n^2+2n} - n)}{(n^2+2n) - (n^2)} = \sqrt{n^2+2n} - n$$
Hence, LHS = RHS.
Note: The statement is not true for negative integers, since the first inequality doesn't hold.
A: Note that $(n+1)^2=n^2+2n+1$, so $\sqrt{n^2+2n}=\sqrt{(n+1)^2-1}$, and therefore
$$\left\lfloor\sqrt{n^2+2n}\right\rfloor=n\;.$$
Thus, 
$$\begin{align*}
\sqrt{n^2+2n}-\left\lfloor\sqrt{n^2+2n}\right\rfloor&=\sqrt{n^2+2n}-n\\
&=\sqrt{n^2\left(1+\frac2n\right)}-n\\
&=n\left(\sqrt{1+\frac2n}-1\right)\\
&=n\left(\sqrt{1+\frac2n}-1\right)\cdot\frac{\sqrt{1+\frac2n}+1}{\sqrt{1+\frac2n}+1}\\
&=\frac{n\left(1+\frac2n-1^2\right)}{\sqrt{1+\frac2n}+1}\\
&=\frac2{\sqrt{1+\frac2n}+1}\;.
\end{align*}$$
Added: I actually verified the result exactly as I’ve written this out, but I’ve a lot of experience. After noticing that the lefthand side is $\sqrt{n^2+2n}-n$, you would probably do better to try to simplify the righthand side. There are two natural ways to begin to do that. One is to multiply it by
$$\frac{\sqrt{1+\frac2n}-1}{\sqrt{1+\frac2n}-1}$$
to rationalize it, and the other is to multiply it by $\dfrac{n}n$ to get rid of the fraction inside the square root. If you rationalize, you’ll still want to multiply by $\dfrac{n}n$ to get rid of the fraction inside the square root; if you get rid of that fraction first, you’ll still want to rationalize, though the rationalizing factor will be different from the one that you use if you rationalize first.
