If I can't use Alt. Series Conv. test, how do I determine if it converges or diverges? Stumped on this absolute convergence problem!
(Converge conditionally, absolutely, or diverges)

$$\sum_{n=1}^\infty \frac{(-1)^n n^2}{(n+3)^2} $$

First, I tested for absolute convergence, and the fraction reduced to $\frac{n^2}{n^2}=1$   .....So, it does not converge absolutely.

Next, to use the alternating series test on the original series, you need to make sure the magnitude of the terms are decreasing.  Well, they are not:

$$\sum_{n=1}^\infty \frac{(-1)^n n^2}{(n+3)^2} = -\frac{1}{16} + \frac{4}{25} - \frac{9}{36} + .....$$

So, without using the alternating series test, how do I determine if the orig. series converges or diverges?
$$\frac{n^2}{(n+3)^2}\xrightarrow[n\to\infty]{}1\implies (-1)^n\frac{n^2}{(n+3)^2}\rlap{\;\;\;\;/}\xrightarrow[n\to\infty]{}0$$
Indeed, Clement's comment below is correct.  This sum is not convergent and has no well-defined value!

The method below is akin to resummation but still very problematic.  For this sum, the Cesaro sum converges to 0 and Abel's resummation method gives $\frac{1}{8} \left(-26+6 \pi ^2-48 \log (2)\right)\approx -0.00667$.  Both differ from the value computed below.

-- I'm leaving the equation below as a reference for Clement's comment, Beware --

One approach is to combine odd and even terms.  This gives a sequence with positive terms, for which convergence is easy to determine.  Thus, we have
$$ \sum_{n=1}^\infty \frac{(-1)^n n^2}{(n+3)^2} = \sum_{k=0}^\infty \left( -\frac{(2k+1)^2}{(2k+4)^2} + \frac{(2k+2)^2}{(2k+5)^2} \right) = \sum_{k=0}^\infty \frac{39+72k+24 k^2}{4(k+2)^2(2k+5)^2} = \frac{3 \pi^2 - 12 \ln (4)-11}{4} \approx 0.493 $$