Is $\cos(\pi \sqrt n)u[n]$ stable?

Question

I took a z transform and got a double pole at $z=1$, but I don't know if that's correct.

I'm lost because I don't know if $\cos(\theta)$ converges or diverges or what that means for $h[n]$ being absolutely summable.

Answer 1

The system with impulse response given by $h[n] = \cos(\pi\sqrt{n})u[n]$ is BIBO-unstable because the sum $\sum_{n=-\infty}^\infty |h[n]]$ diverges instead of being convergent as is needed for BIBO-stability. Note that for all positive integers $k$, $h[k^2]=\cos(\pi k)$ has value $\pm 1$ and so $$\sum_{n=-\infty}^\infty |h[n]| = \sum_{n=0}^\infty |\cos((\pi\sqrt{n})|$$ is a sum that contains infinitely many $+1$ terms (and all the other terms are guaranteed to be positive too since $\cos(\pi r)=0$ if and only if $r = k+\frac 12$ where $k$ is an integer, and there is no integer $n$ whose square root is of the form $k+\frac 12$). So, $\sum_{n=-\infty}^\infty |h[n]]$ diverges, and the system is BIBO-unstable.

Answer 2

The LTI system defined by the impulse response $$h[n] = \cos(\pi \sqrt{n} ) u[n] $$ is unstable, as the absolute sum of the impulse response does not converge and diverges to infinity instead, i.e.;

$$ \sum_{n=-\infty}^{\infty} |h[n]| = \sum_{n=0}^{\infty} |\cos(\pi \sqrt{n}) u[n]| \longrightarrow \infty $$

Answer 3

The summation of $\cos(\pi (z^{n^(0.5)})) z^{-n}$ should converge for the $z$ transform to converge and it is stable if the ROC includes the unit circle. Since you got double poles at $|z|=1$, the causal ROC will lie outside the largest pole and hence it won't include $|z|=1$ and hence it cannot be stable.