Please help me to understand a proof that $a_n = \frac{\tan 1}{2} + \frac{\tan2 }{2^2} + \dots + \frac{\tan n}{2^n}$ is a Cauchy sequence

Question

$\textbf{Problem:}$ Show that $$a_n = \frac{\tan 1}{2} + \frac{\tan2 }{2^2} + \dots + \frac{\tan n}{2^n}$$ is a Cauchy sequence.

This is the solution.

Here's my question: In the first line, we have $\arctan$ or $\tan$? $a_n$ is $\tan$ or $\arctan$? I mean the solution is wrong?

Answer 1

I'm quite sure that the question is wrong and that whoever asked it meant to write $\arctan$ instead of $\tan$.

Answer 2

The fact that $\sum_{n\geq 1}\frac{\arctan n}{2^n}$ is convergent is fairly trivial, the fact that $$\sum_{n\geq 1}\frac{\tan n}{2^n} $$ is convergent... not so much. We may recall that $\pi$ has a finite irrationality measure, hence $$ \left|\frac{\pi}{2}-\frac{p}{q}\right|\leq \frac{1}{q^{10}} $$ occurs for a finite number of $\frac{p}{q}\in\mathbb{Q}$. In particular even if $\left|\tan n\right|$ is large, since $n$ is close to an odd, integer multiple of $\frac{\pi}{2}$, it cannot be larger than, say, $n^{12}$, with a finite number of exceptions. Since $\sum_{n\geq 1}\frac{n^{12}}{2^n}$ is convergent, $\sum_{n\geq 1}\frac{\tan n}{2^n}$ is convergent.

But dealing with $\tan$ (in place of $\arctan$) requires to invoke some fact from Diophantine approximation.