I know that the $\lim_{x\rightarrow\infty}\tan(x)$ does not exist. Now my question: Is the sequence $\tan(n)$ convergent for $n\in\mathbb N$?
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$\begingroup$I claim that $\tan(n)$ is not convergent. Assume to contrary $\lim \tan(n)=l$.
We can use the following trigonometry identity $\tan(n+1)=\frac{\tan(n)+\tan(1)}{1-\tan(n)\tan(1)}$. Thus we have $l=\frac{l+s}{1-ls}$ where $\tan(1)=s$ and so $l-l^2s=l+s$. It is not hard to see that $s(1+l^2)=0$ and so a contradiction.
$\endgroup$ 0 $\begingroup$The period of $\tan x$ is $\pi$. Note $\tan x$ has undefined values at $\pi/2 + k\pi$, zero values at $k\pi$ for $k \in \mathbb{Z}$, and alternates in sign between these integer multiples of $\pi/2$. Since $1 \lt \pi/2$, and $\pi/2$ is irrational, the sequence $\{\tan(n)\}$ will change signs infinitely often (because the step size of $1$ is too small for us to leave negative values of $\tan n$ without hitting positive values ahead, and conversely).
In such a case the only limit possible is $0$. But this is easily ruled out as well, since if $\tan n$ is close to zero, then $n$ is close to some $k\pi$. Then, looking at the sum angle formula for $\tan$ cited by @BabakMiraftab, $\tan(n+1)$ will be close to $\tan 1$ and $\tan(n-1)$ will be close to $-\tan 1$. Since $\tan 1$ is not zero, there is no limit.
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