True or False (Calculus 3 question)?

By the ratio test, the series converges if abs((a(k+1)/a(k)) * x) < 1, where abs

indicates absolute value. Since it converges fro x = 8, we know that

abs(a(k+1)/a(k)) < (1/8), and thus abs((a(k+1)/a(k))*(-7)) < 1, and so the series

converges for x = -7.

Edit: If the series were sum(a(k) * (x - 7.5)^k) then your concern would be valid, and

the statement would be false. (As a counterexample look at a(k) = (-1)^k * (1.5)^k.)

But because the series is expressed with just x^k, the midpoint of the interval

of convergence must be 0, and hence if it converges for x = 8 it must converge

for any x such that lxl < 8.

Edit: I made my last edit before I saw your last edit, so now I'll respond to that.

I don't think this works because the reason sum(a(k) * 7^k) converges might

be because it is an alternating series, i.e., there may be a (-1)^k factor in a(k),

in which case if we stuck in (-7)^k the series would no longer be alternating

and so we wouldn't be able to make any conclusions about convergence

at x = -7. This doesn't invalidate the statement but it doesn't prove it, either.

The critical elements are that we have the term x^k in the series and that it

converges for x = 8.

true...radius of convergence is at least 8 ;