A Quick, Calculus II-Level Proof Regarding the Divergence of Certain Series? I came across the following problem in the Calculus text I'm using to teach my class:
Suppose $\sum a_n$ is a positive series such that $\{a_n\}$ is a decreasing sequence and the sequence $\{na_n\}$ converges, but not to zero. Show that $\sum a_n$ diverges.
Now, this is a fairly standard analysis problem that can be shown using, for example, the Cauchy Condesation Test, among other things, but my students do not have access to this, nor any of the other standard analytic arguments one would use. The hint given in the text is to use the Limit Comparison Test to compare $\sum a_n$ with an "appropriate" series, but I have not been able to figure out what the authors intended.
Secondly, the students are supposed to use this fact to provide a "quick proof" that
$$\sum \frac{\arctan n}{\sqrt{n}}$$
diverges. It's obvious this is a positive series, but I don't see a "quick" Calculus-level argument that it's sequence of terms is decreasing. Furthermore, the sequence $\{na_n\} =: \{\sqrt{n} \arctan n \}$ does not converge, so I don't see how the above fact applies directly. 
In each case, what is the text is looking for that the average Calculus II student is supposed to see?
EDIT: The main thing throwing me off here was that students were supposed to somehow "quickly" show that $\frac{\arctan n}{\sqrt{n}}$ is decreasing (for example, in order that solving this problem with this fact is justified, the proof should be quicker than direct use of the Limit Comparison Test, which it is not.)However, as pointed out below, the sequence $\{a_n\}$ actually does NOT need to be decreasing. That makes the problem trivial!

If $na_n\to c\ne 0$ use limit comparison with $b_n = c/a_n$. (By the way, the decreasing hypothesis is not needed.) 
For the second one, direct limit comparison is obvious. I’m not sure what they had in mind. 
