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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.

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Is $n$ an integer? –  Calvin Lin Jan 8 '13 at 15:04
    
Yes (more text) –  Krau Jan 8 '13 at 15:04
    
You will need the condition that $n$ is a non-negative integer. –  Calvin Lin Jan 8 '13 at 15:11
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2 Answers

up vote 7 down vote accepted

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.

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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.

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