Answer
$p$ is not a negative integer.
Work Step by Step
Case 1: When $p\geq 0$
$\lim\limits_{n \to \infty}\dfrac{1}{n+p}=0$ .
This implies that the limit $0$ satisfies all the conditions for the alternating series test and so, the series converges by the Test of Divergence.
Case 2: When $p \lt 0$
Here, the limit for $\lim\limits_{n \to \infty}\dfrac{1}{n+p}$ is undefined because $n=-p$ and so, the series does not converge by the Test of Divergence.
However, for the negative values, the denominator does not become $0$.
Hence, the value of $p$ can have any real value but $p$ is not a negative integer.
