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


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!

  • $\begingroup$ Your asymptotic equivalence is off by a factor of $2$. $\endgroup$
    – joriki
    Aug 1, 2018 at 10:35
  • $\begingroup$ What (in your view) is the difference between showing that the limit of the ratio is $2$ and showing that $$\frac2{\sqrt{n}+2}\sim\frac2{\sqrt{n}}\;?$$ $\endgroup$
    – joriki
    Aug 1, 2018 at 10:36
  • $\begingroup$ @joriki nothing, so the processes are the same. Thank you! $\endgroup$ Aug 1, 2018 at 10:39
  • $\begingroup$ You can also use $\frac{C}{n}$ as a lower bound. $\endgroup$
    – Diger
    Aug 1, 2018 at 10:40
  • 2
    $\begingroup$ The problem is how much do your readers need convincing. For a text aimed to people that know this you wouldn't even need to stop to make the observation that the series is equivalent to $\sum2/\sqrt{n}$. For an instructor that is evaluating that you know the limit form of the comparison test, you would need to show them that you know it. If you were writing for students that have learned even less, then even writing the limit test wouldn't be enough. You would need to prove that since the limit exists and is $\neq0$, that implies an inequality that makes the series diverge. $\endgroup$
    – user580788
    Aug 1, 2018 at 10:41

2 Answers 2


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

  • $\begingroup$ This is a direct comparison test, right? If this didn't come to mind immediately, is my 'limit' comparison argument enough. $\endgroup$ Aug 1, 2018 at 10:36
  • $\begingroup$ Yes this is a comparison test, since $$\frac{1}{\sqrt{n}}$$ is divergent. $\endgroup$ Aug 1, 2018 at 10:38
  • 1
    $\begingroup$ It gives $$\sqrt{n}\geq 2$$ so $$n\geq 4$$ that is a enough. $\endgroup$ Aug 1, 2018 at 10:43

Let $n \gt 4$:

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

Limit comparison test, harmonic series.


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