# Is this a power signal?

$x[n]=\sin(\pi n)$

Since this is a discrete signal, and n can take only values like ...,1,2,3 ... the value of the power is getting zero, and since it gets zero does it mean that it's not a power signal?

There a few variations on the definitions of "energy signal" and "power signal". For example, Sklar in his textook "Digital Communications" defines "energy signal" as a signal with energy $E$ such that $0 < E < \infty$, and a "power signal" as a signal with power $P$ such that $0 < P <\infty$.
However, others (see for example this answer) define "energy signal" as a signal with $E < \infty$.
In your case, $x[n]=0$ for all $n$; then this signal has zero energy and zero power. Whether it is classified as an energy or a power signal depends on which specific definition you adopt. For instance, if you go with Sklar's, then this signal is neither an energy signal nor a power signal.
• Well, it is both an energy signal (with zero energy) and a power signal (with zero power). Note that $\frac{1}{2N+1}\sum_{n=-N}^N |x[n]|^2$ "approaches" a finite limit as $N \to \infty$ and so the signal is a power signal. That the average is already at the limit $0$ when $N=0$ and just stays there forever is not germane. – Dilip Sarwate Feb 25 '18 at 3:34