Question

1) If ∑Cn(x–3)^n converges at x=7 and diverges at x=10, what can you say about the convergence at x=11? At x=5? At x=0?

Answer 1

The given sequence converges at x=7 and diverges at x=10. It is probable that the sequence converges at x=11 and x=5, but unlikely to converge at x=0.

Step-by-step explanation:

The given sequence, ∑Cn(x–3)^n, converges at x=7 and diverges at x=10. To determine the convergence at x=11, x=5, and x=0, we need to analyze the behavior of the sequence near these values. Since the sequence converges at x=7, it is likely to converge for nearby values as well. Therefore, it is probable that the sequence converges at x=11 and x=5. However, as the sequence diverges at x=10, it is unlikely to converge at x=0.

Answer 2

The power series converges at x=7 and diverges at x=10, indicating a radius of convergence of at most 3. Therefore, it will diverge at x=11, which is 4 units away from the center of convergence, and converge at x=5, which is within the radius. It will also diverge at x=0, which is outside the radius.

Step-by-step explanation:

If ∑Cn(x−3)n converges at x=7 and diverges at x=10, we can determine the convergence at other points based on the radius of convergence of the power series. The distance between the two known points of convergence and divergence (7 and 10) is 3 units, which means the radius of convergence is at most 3. Therefore, since x=11 is 4 units away from 7 (beyond the radius of convergence), the series will diverge. Conversely, x=5 is only 2 units away, and the series will converge at this point. At x=0, which is 7 units away, the series will diverge as well.