Convergence of the series $\sum_{n=1}^{\infty}\cot^{-1}(n)$ Consider the following series
 $$\sum_{n=1}^{\infty}\cot^{-1}(n)$$ 
Which test can be useful to show convergence or divergence of above series?
I tried the comparison and Ratio test, but they did not help me. 
 A: The integral test works.  To find $\int \cot^{-1}x\,dx$, let $u = \cot^{-1} x$ and $dv = dx$.  Then $du = \frac{-1}{1+x^2}\,dx$ and $v=x$, so:
$$
    \int \cot^{-1}x\,dx = x \cot^{-1} x + \int \frac{x}{1 + x^2}\,dx
    = x \cot^{-1} x + \frac{1}{2}\ln(1+x^2) + C
$$
Therefore
$$
    \int_0^\infty \cot^{-1}x\,dx
    =\lim_{t\to\infty} \left(t \cot^{-1}t + \frac{1}{2}\ln(1+t^2)\right)
$$
(when we evaluate at $t=0$, both $t \cot^{-1} t$ and $\frac{1}{2}\ln(1+t^2)$ are zero.)
By L'Hôpital's rule:
\begin{align*}
    \lim_{t\to\infty} t \cot^{-1} t 
    &= \lim_{t\to\infty} \frac{\cot^{-1}{t}}{1/t} \\
    &= \lim_{t\to\infty} \frac{-1/(1+t^2)}{-1/t^2} 
     = \lim_{t\to\infty} \frac{t^2}{1+t^2} = 1
\end{align*}
But as the other term, $\lim_{t\to\infty}\frac{1}{2}\ln(1+t^2) = +\infty$.
Therefore the integral diverges, which means the series diverges as well.

Although the integral test is definitive, you expend a lot of effort where a little intuition can save it.  So here is some intuition: remember that for small $t$, $\sin t \approx t$ and $\cos t\approx 1$.  This means that 
$$
    \cot t \approx \frac{1}{t} 
$$
and therefore that
$$
     \cot^{-1}\frac{1}{t} \approx t
$$
(again, for sufficiently small $t$).
If $n$ is large, $t = \frac{1}{n}$ is small, so
$$
    \cot^{-1} n \approx \frac{1}{n}
$$
Since we know that $\sum_{n=1}^\infty \frac{1}{n}$ diverges, we have a good clue that $\sum_{n=1}^\infty \cot^{-1} n$ diverges too.  We can confirm that with the limit comparison test: 
$$
    \lim_{n\to\infty} \frac{\cot^{-1} n}{1/n} = 1
$$
as we showed already.
