TITLE: Comparing to a constant with the limit comparison test
QUESTION [1 upvotes]: Recently in my calculus class, we covered the limit comparison test for infinite series (among other things related to sequences and series). We have also covered improper integrals earlier this semester. One of the homework questions asks us to "calculate" $$\sum_{n=1}^\infty a_n\:\:\text{where}\:\:a_n=\arctan\left(\frac {n^2}{n^2+4}\right).$$
I noted that, as $a_n\to \infty$, $\arctan\left(\frac {n^2}{n^2+4}\right)\approx \arctan(1)$, and used that to compare using the limit comparison test, resulting in $$\lim_\limits{n\to\infty}\frac{\arctan\left(\frac {n^2}{n^2+4}\right)}{\arctan(1)} = 1,$$ so I concluded that $a_n$ diverges, since $$\sum_{n=1}^\infty \arctan(1) = \infty.$$ 
I was wondering if, both in this specific case and in general, using a constant with the limit comparison test was even allowed, since up until now we have been comparing to similar $f(x)$ or $b_n$ for integrals and series respectively.

REPLY [2 votes]: Yes: the function $f(x) = \arctan 1$ (independent of $x$) is a perfectly valid function.
Note that in such cases, the Test for Divergence will work more quickly, since the summand does not tend to $0$.
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