Is the signal
$ \ x(t) = \sum_{k = - \infty}^{\infty} \frac{1}{(t + 2k)^{2} + 1} $
periodic? Answer yes/no and justify your answer mathematically.
Yes, because $ x(t + 2) = \sum_{k = - \infty}^{\infty} \frac{1}{(t + 2 + 2k)^{2} + 1} = \sum_{k = - \infty}^{\infty} \frac{1}{(t + 2 (k + 1))^{2} + 1} $
let $ \ r = k + 1, \sum_{k = - \infty}^{\infty} \frac{1}{(t + 2r)^{2} + 1} = x(t) $
