Convergence of $\sum^{\infty}_{n=1}\frac{2}{\sqrt{n}+2}$

Question

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}. $

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 $p=\frac{1}{2}$):

$$\sum^{\infty}_{n=1}\frac{2}{\sqrt{n}+2}\space\text{diverges}$$

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!

Answer 1

Note that $$\frac{2}{\sqrt{n}+2}\geq \frac{1}{\sqrt{n}}$$

Answer 2

Let $n \gt 4$:

$\dfrac{2}{2+√n} \gt \dfrac{2}{2√n}\gt \dfrac{1}{n}.$

Limit comparison test, harmonic series.