Let's write #sum_(n=0)^(oo) (-1)^n * x^(n+8)# in the form of #sum_(n=0)^(oo) a_n x^n#.
#sum_(n=0)^(oo) (-1)^n * x^(n+8) = sum_(m=0)^(oo) a_m x^m#,
where #a_m = 0 if m<8 and a_m = (-1)^(m-8) if m>=8#.
Now you can compute # R = 1/(lim "sup"_(m->oo)root(m)(abs(a_m)))#.
Yet, #abs(a_m)# can only take #0# or #1# as a value. But when #m# tends to infinity (so when #m# is much larger than #8#), #abs(a_m) = 1 = root(m)(abs(a_m)# so #lim "sup"_(m->oo)root(m)(abs(a_m)) = 1#.
Therefore, #R = 1/1 = 1#.
You now know that #sum_(n=0)^(oo) (-1)^n * x^(n+8)# converges for #x in ]-1, 1[#.
What about #x = -1# or #x = 1# ?
If #x = -1# :
#sum_(n=0)^(oo) (-1)^n * (-1)^(n+8) = sum_(n=0)^(oo) (-1)^(2n + 8) = sum_(n=0)^(oo) 1 = +oo#
If #x = 1# :
#sum_(n=0)^(oo) (-1)^n * (1)^(n+8) = sum_(n=0)^(oo) (-1)^n# is not defined.
So #sum_(n=0)^(oo) (-1)^n * x^(n+8)# converges for #x in ]-1, 1[#.
