In the Comparison Tests section of my textbook, I am tasked with determining the convergence of the series $$\sum^{\infty}\_{n=1}\frac{2}{\sqrt{n}+2}$$

I will argue the series diverges using the "Limit" Comparison test. 

Consider $$a\_n=\frac{2}{\sqrt{n}+2}$$ and $$b\_n=\frac{1}{\sqrt{n}}$$ Note: The latter series $b\_n$ is a divergent $p$-series with $p=\frac{1}{2}. 
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If $$\lim\_{n\rightarrow\infty}\frac{a\_n}{b\_n}=c$$ where $c>0$, then either both $a\_n$ and $b\_n$ converge, or both $a\_n$ and $b\_n$ diverge. 

Thus $$\lim\_{n\rightarrow\infty}\frac{a\_n}{b\_n}=\lim\_{n\rightarrow\infty}\frac{2\sqrt{n}}{\sqrt{n}+2}=\lim\_{n\rightarrow\infty}\frac{2}{1+\frac{2}{\sqrt{n}}}=2$$

Since the above limit existed as a finite value $c=2>0$ and since $\sum b\_n$ diverges ($p$-series with $p=\frac{1}{2}$):

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> $$\sum^{\infty}\_{n=1}\frac{2}{\sqrt{n}+2}\space\text{diverges}$$
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However, from the beginning I can see the asymptotic equivalence: $$\frac{2}{\sqrt{n}+2}\sim\frac{1}{\sqrt{n}}$$ Why do I need the limit comparison test when I can deduce the result quickly upon inspection? Thanks in advance!