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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.

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    The integral test would work. – 2017-02-28
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    "I tried the comparison ... test" Good, how? Please be specific. – 2017-02-28
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    @joeb I'm pretty sure the OP means $\operatorname{arccot}$. Writing the series in terms of $\arctan$ would probably help. – 2017-02-28
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    I know that the value of $\cot^{-1}(x)$ is between 0 and $\pi$, so it is greater than $\frac{-1}{x}$ to compare. for posetive $x$ – 2017-02-28

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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.