## Task 27/64

Determine the sequence $\left(x_{n}\right)$ of those $x_{n}$-values for which the following holds: $\sin \frac{1}{x_{n}}=1$.
Due to the periodicity of the sine function, the validity of the equation

$$
\sin \frac{1}{x_{n}}=1 \quad \text { implies the equation } \quad \sin (a+2 k \pi)=1
$$

($k$ arbitrary, integer) with $\frac{1}{x_{n}}=a+2 k \pi$ or $x_{n}=\frac{1}{a+2 k \pi}$. Since $\sin (a+2 k \pi)=\sin a=1$ implies $a=\frac{\pi}{2}$, it follows that

$$
x_{n}=\frac{2}{\pi+4 n \pi}=\frac{2}{\pi(1+4 n)}
$$

That the sequence

$$
\left(x_{n}\right)=\left(\frac{2}{\pi(1+4 n)}\right)
$$

whose first terms are $x_{0}=\frac{2}{\pi}, x_{1}=\frac{2}{5 \pi}, x_{-1}=\frac{2}{-3 \pi}, x_{2}=\frac{2}{9 \pi}, x_{-2}=\frac{2}{-7 \pi}, \ldots$, is a null sequence, is evident without further ado.