series converges or diverges?

$\sum_{k\ =\ 0}^n [k(k+2)]/(k+3)^2$

the question says to determine if the series is convergent or divergent, & if it's convergent, to evaluate it. i thought maybe i should multiply it out, so i did, and i got

$\sum_{k\ =\ 0}^n (k^2+2k)/(k^2+6k+9)$

i'm not sure how much this helps, though.
the only tests that we are "supposed" to know for this section are the test for divergence (that says that if the limit of the sequence DNE or is not zero then the sum is divergent) and the theorem that says that if the limit is zero, it is convergent.

i tried to do this limit but i was a bit confused..can someone help?

edit:
i tried doing the limit, and got it to equal 1
is that correct? :/

yes, dividing through by k^2 does in fact give you 1 ..So I'm assuming the series does converge to 1 but then again its been a long time.

well, i thought that if the limit equals anything but zero, it diverges, but i'm not 100% sure..

$\lim_{k\rightarrow +\infty} \frac{k(k+2)}{(k+3)^2}$ is not 0 (it is 1) therefore the series diverges