Answer
Converges to $1$
Work Step by Step
Consider $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \tan hn $
Since, $\lim\limits_{n \to \infty}e^n=\infty$
So, $\lim\limits_{n \to \infty} \tan hn =\lim\limits_{n \to \infty} \dfrac{e^n-e^{-n}}{e^n+e^{-n}}=\lim\limits_{n \to \infty} \dfrac{1-1/e^{2n}}{1 +1/e^{2n}} $
or, $=1$
Hence, $\lim\limits_{n \to \infty} a_n=1$ and {$a_n$} is convergent and converges to $1$
