Q: Is the signal: $ x(t)=\sum_{k=-\infty}^\infty k = \frac{1}{(t+2k)^2+1} $ periodic?
A: Yes:
$ x(t+2)=\sum_{k=-\infty}^\infty k \frac{1}{(t+2+2k)^2+1} $
$ =\sum_{k=-\infty}^\infty k \frac{1}{(t+2+2k)^2+1} $
$ =\sum_{k=-\infty}^\infty k \frac{1}{(t+2(k+1))^2+1} $ Setting $ r = k+1 $
$ =\sum_{k=-\infty}^\infty k = \frac{1}{(t+2r)^2+1} $
$ =\text{x(t+2)} $
