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[.
