The series looks like: 1, 1/7, 0, -1/2197, 1/4096, ...
an+1=(-n-1+2)n+1/(3n+3+4)n+1=(-n+1)n+1/(3n+7)n+1; an=(-n+2)n/(3n+4)n;
an+1/an=(-n+1)n+1(3n+4)n/[(-n+2)n(3n+7)n+1].
an+1/an→ as (-n)n+1(3n)n/[(-n)n(3n)n+1]=-n/3n=-⅓ as n→∞. The value of an+1/an is less than 1 for n, suggesting convergence. Since the ratio of two consecutive term roughly follows the rule an+1=-an/3, then the sum of two consecutive terms approximates to an-an/3=⅔an, and if we pair consecutive terms such that this sum is positive, i.e., a2k>a2k+1 for even n, represented by n=2k. These pairs make up a series of positive terms in which the consecutive term ratio is about ⅔, which is less than 1, implying absolute convergence. Note that the ratio test doesn't work when n=2.
(In fact the sum of the terms converges to about 1.14257.)