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The sequence in question is $$ a_n = \sqrt{n^2+4}-n\,. $$ I can see that it is strictly decreasing by finding the derivative and observing that it is negative on the entire range containing the relevant values of $n$. But I feel like there must be a simpler algebraic reason that I'm just not seeing.

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    $\begingroup$ One hand-wavy reason why this is decreasing is that the expression is measuring how much the “$+4” matters to the square root. As $n$ grows larger, the “+4” matters less and less, so the sequence decreases. $\endgroup$ May 13, 2018 at 15:36

4 Answers 4

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Notice that we have $$\sqrt{n^2+4}-n = \frac{4}{\sqrt{n^2+4}+n}$$

hence as $n$ increases, the expression decreases as the denominator is positive and increases while the numerator is a positive constant.

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Alternatively, you can show: $$ \begin{split} a_n>a_{n+1} & \iff \sqrt{n^2+4}-n>\sqrt{(n+1)^2+4}-(n+1)\\ & \iff \sqrt{n^2+4}+1>\sqrt{n^2+2n+5}\\ & \iff n^2+4+2\sqrt{n^2+4}+1>n^2+2n+5\\ & \iff \sqrt{n^2+4}>n\\ & \iff n^2+4>n^2\\ & \iff 4>0 \end{split} $$

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Write it as $a_n = \sqrt{n^2 + 4} - \sqrt{n^2}$.

Interpret it as "the increase in $\sqrt{x}$ when $x$ is increased by $4$", for $x = n^2$. Note that this is decreasing in $x$ iff it's decreasing in $n$, since all these numbers are positive.

Another way of thinking about this is as "the amount by which I need to increase $y = \sqrt{x}$ to make $y^2$ increase by $4$".

Since $y^2$ grows faster for large $y$, this amount is going to get smaller for large $y$.

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[EDIT] I just saw that you tried this method. I will not delete this post as it might help other users.

Another approach that generally works:

Let $f: \mathbb{R} \to \mathbb{R}: n \mapsto \sqrt{n^2 + 4} - n$

Then

$$f'(n) = \frac{n}{\sqrt{n^2 + 4}}-1 = \frac{n}{n \sqrt{1 + 4/n^2}} - 1 = \underbrace{\frac{1}{\sqrt{1+4/n^2}}}_{>1}-1 < 0$$

Hence the function is decreasing on $\mathbb{R}$ and surely on $\mathbb{N}$.

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