Convergence of a series involving arccot? I am supposed to determine the convergence (conditional or absolute) or the divergence of the series
$$\sum_{n=1}^\infty \frac{\cot^{-1} ({\pi/4 + n\pi})}{\sqrt{3n^3+1}}$$
I previously solved a similar problem but the function in question was not $\cot^{-1} ({\pi/4 + n\pi})$  but rather $n\tan^{-1} ({\pi/4 + n\pi})$ and that problem was far easier to solve because I concluded that when $n\to+\infty$, $\tan^{-1} ({\pi/4 + n\pi}) \to\pi/2$. Then I used the Limit Comparison Test with the sequence $\frac{1}{n^2}$, and therefore I concluded that the given series absolutely converges. I believe it is important to keep in mind that the denominator in the other problem was different.
My problem with this series is that from what I could find on the internet $\cot^{-1} ({\pi/4 + n\pi}) \to 0$ as $n\to+\infty$, but by using this line of thinking the whole series will look something like
$$\sum_{n=1}^\infty \frac{0}{\sqrt{3n^3+1}}$$ which makes no sense at all.
Am I looking at these problems wrong, or is there another value to $\cot^{-1} ({\pi/4 + n\pi})$ as $n\to+\infty$?
 A: Note that
$\cot(x)
=\dfrac1{\tan(x)}
$,
so that if
$\tan(x_n)
\to L
$,
then
$\cot(x_n)
\to \dfrac1{L}
$.
Also note that
$\arctan(x)+\text{arccot}(x)
=\pi/2
$.
Therefore
if
$y_n \to \infty$
(as $\pi/4+n\pi$ does),
then
$\arctan(y_n)
\to \pi/2
$
and
$\text{arccot}(y_n)
\to 0
$.
So your result is correct.
Another thing to note:
The whole 
"$\pi/4+n\pi$"
bit is nonsense,
designed to fool you.
Taking the
tan or cot of
"$\pi/4+n\pi$"
might be interesting,
but the inverse tan or cot of this
is just the inverse tan or cot
of a large number,
nothing more.
A: As n gets large $cot^{-1}(n) \to 1/n.$
and you can run a limit comparison test of 
$\frac{\cot^{-1} ({\pi/4 + n\pi})}{\sqrt{3n^3+1}}$vs.$\frac{1}{n\sqrt{3n^3+1}}$

The Limit comparison test.
If you have two series $\sum a_n, \sum b_n$ and
$\lim_\limits{n\to\infty} |\frac{a_n}{b_n}| = M$ with $M>0$ and bounded. Then either both series converge or both series diverge.
In this case, with $a_n, b_n$ as above: $\frac{a_n}{b_n} = n\cot^{-1}(\pi/4+n\pi)$.
$\lim_\limits{n\to\infty} n\cot^{-1}(\pi/4+n\pi) = \frac{1}{\pi}$.
Now, can you show that
$\sum\frac{1}{n\sqrt{3n^3+1}}$ converges?
