Why is the following sequence monotonically decreasing? 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.
 A: 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 $y$.
A: 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.
A: [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 $\mathbb{N}$.
A: Alternatively, you can show:
$$
\begin{split}
a_n>a_{n+1} 
& \iff \sqrt{n^2+4}-n>\sqrt{(n+1)^2+4}-(n+1)\\
& \iff \sqrt{n^2+4}+1>\sqrt{n^2+2n+5}\\
& \iff n^2+4+2\sqrt{n^2+4}+1>n^2+2n+5\\
& \iff \sqrt{n^2+4}>n\\
& \iff n^2+4>n^2\\
& \iff 4>0
\end{split}
$$
