Why is the following sequence monotonically decreasing?

Solution 1:

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.

Solution 2:

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

Solution 3:

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 $y$.

Solution 4:

[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 $\mathbb{N}$.