If you have a alternating series you and put it though the nth term test you split the limit into two parts multiplying each other. One containing (-1)^n and another containing what its multiplied by. But the limit containing (-1)^n will always result in DNE(Does not exist) so if you multiply that by a number whatever the other limit answer is. Would that still be DNE? And would mean that the divergence test(nth term test) always passes with the alternating series since it does not equal 0 and always equals DNE. DNE times number
I saw some other questions on stack exchange similar but they did not answer my specific question about DNE so that is why I am asking.
Your method of taking limits is not good: you can't legitimately drag "nonsense/does not exist" limits through calculations whenever you want. If you have a sequence $a_n = b_nc_n$ where $\lim_{n \to \infty} b_n$ does not exist while $\lim_{n \to \infty} c_n$ exits, that does not mean $\lim_{n \to \infty}$ does not exist.
Example. Let $a_n = (-1)^n/n^2$, with $b_n = (-1)^n$ and $c_n = 1/n^2$. Then $\lim_{n \to \infty} b_n$ does not exist, $\lim_{n \to \infty} c_n = 0$, and $\lim_{n \to \infty} a_n = 0$.
Example. Let $a_n = 1/n = n/n^2$, with $b_n = n$ and $c_n = 1/n^2$. Then $\lim_{n \to \infty} b_n = \infty$, $\lim_{n \to \infty} c_n = 0$, and $\lim_{n \to \infty} a_n = 0$.
Example. Let $a_n = 1 = n/n$, with $b_n = n$ and $c_n = 1/n$. Then $\lim_{n \to \infty} b_n = \infty$, $\lim_{n \to \infty} c_n = 0$, and $\lim_{n \to \infty} a_n = 1$.