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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
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



$$\begin{align*} \sqrt{n^2+2n}-\left\lfloor\sqrt{n^2+2n}\right\rfloor&=\sqrt{n^2+2n}-n\\ &=\sqrt{n^2\left(1+\frac2n\right)}-n\\ &=n\left(\sqrt{1+\frac2n}-1\right)\\ &=n\left(\sqrt{1+\frac2n}-1\right)\cdot\frac{\sqrt{1+\frac2n}+1}{\sqrt{1+\frac2n}+1}\\ &=\frac{n\left(1+\frac2n-1^2\right)}{\sqrt{1+\frac2n}+1}\\ &=\frac2{\sqrt{1+\frac2n}+1}\;. \end{align*}$$

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


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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