Question

For \(a_n \in \mathbb{C}\), \(n = 0,1,2,\ldots\), the power series \[ \sum_{n=0}^{\infty} a_n (z-2)^n \] converges at \(z = 5\) and diverges at \(z = -1\). Then the radius of convergence of this power series is equal to ______. (Answer in integer)

Hint:

Both given points are the same distance from the center of the series, so the radius of convergence must equal that common distance.

Answer 1

Correct Answer: 3

Step 1: Recall the basic fact about radius of convergence.
For a power series \(\sum_{n=0}^{\infty} a_n (z-z_0)^n\) with radius of convergence \(R\), the series converges for every \(z\) with \(|z-z_0| < R\), and it diverges for every \(z\) with \(|z-z_0| > R\). Here the center is \(z_0=2\).

Step 2: Use the convergence information at \(z=5\).
The distance from the center to \(z=5\) is \[ |5-2| = 3. \] Since the series converges at this point, this point cannot lie outside the disc of convergence, so \[ R \geq 3. \]
Step 3: Use the divergence information at \(z=-1\).
The distance from the center to \(z=-1\) is \[ |-1-2| = |-3| = 3. \] Since the series diverges at this point, this point cannot lie inside the disc of convergence, so \[ R \leq 3. \]
Step 4: Combine both bounds.
From Step 2, \(R \geq 3\); from Step 3, \(R \leq 3\). The only value that satisfies both at once is \[ R = 3. \]
Final Answer:
The radius of convergence of the power series is 3. \[ \boxed{R = 3} \]