how to show the following seriers is not convergent How to show the following series does not converge?
$$\sum_{n=1}^{\infty}n^{-1/3} \cot\left( \frac 1 n \right).$$
Thank you very much.
 A: Hint: over $(0,1]$ we have $\cot x\geq\frac{\cot 1}{x}$, hence $\cot\frac{1}{n}\geq n\cdot\cot 1.$
A: Hint: have you tried the limit test?
A: Since $\lim_{x \rightarrow 0} \frac{\tan x }{x} = 1$, we have $\tan x < 2 x$ for sufficiently small positive $x$.
So there is some natural number $N$ such that  $\cot \frac{1}{n}>  {n \over 2}  $ for $n  \ge N$. 
Hence we have $n^{- {1 \over 3 }} \cot ( {1 \over n}) > {n^{2/3} \over 2} $   for $n \ge N$, which implies the divergence.
A: If $n$ is a large number, so $1/n$ is close to $0$, then $\tan(1/n)$ is scarcely distinguishable from $1/n$, so $\cot(1/n)$ is scarcely distinguishable from $n$.
To see this, just draw the circle of unit radius centered at $(0,0)$ in the $(x,y)$-plane, and starting from $(1,0)$, go counterclockwise along the circle a distance of $1/n$.  The tangent of that little angle is the height you have to go up to vertically from $(1,0)$ to the point where you hit that ray that goes out from $(0,0)$.  Going up vertically is nearly the same as going counterclockwise around the circle when you're near that point where the arc of the circle is nearly a vertical line.
So you're looking at $\dfrac{\cot(1/n)}{n^{1/3}}\approx \dfrac n {n^{1/3}}$.
If you want to be more logically precise, just show that $\cot(1/n)>n/2$.  That's a very conservative bound, but it's plenty for the purpose.
