Question 579908
It is not a geometric series.
For odd n, (-1)^n=-1 and [(-1)^n+1]=0, so {{{a[n]=0}}}  
For even n, (-1)^n=1 and  [(-1)^n+1]=2, so {{{a[n]=2/sqrt(n)}}} .
In general, {{{0<= a[n]<=2/sqrt(n)}}} , so {{{a[n]}}} can be made less than any {{{epsilon}}}  by just making {{{n>4/epsilon^2}}}. So it meets the definition of converging to zero.
However, the way you are expected to prove that it converges may be different.