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: Note that both given points are the same distance from the center.
The power series is centered at $z_0=2$ since it is written in powers of $(z-2)$. The point $z=5$ is at distance $|5-2|=3$ from the center, and the point $z=-1$ is at distance $|-1-2|=3$ from the center as well. So both points lie on the same circle $|z-2|=3$, but one gives convergence and the other gives divergence.

Step 2: Suppose, for contradiction, that $R<3$.
If $R$ were less than $3$, then $z=5$, at distance exactly $3>R$ from the center, would lie strictly outside the disc of convergence $|z-2|<R$. A power series always diverges strictly outside its disc of convergence, which contradicts the fact that the series converges at $z=5$. So $R<3$ is impossible.

Step 3: Suppose, for contradiction, that $R>3$.
If $R$ were greater than $3$, then $z=-1$, at distance exactly $3<R$ from the center, would lie strictly inside the disc of convergence $|z-2|<R$. A power series always converges strictly inside its disc of convergence, which contradicts the fact that the series diverges at $z=-1$. So $R>3$ is impossible too.

Step 4: Conclude the only possible value.
Since $R<3$ and $R>3$ are both impossible, the only value left is $R=3$.
\[ \boxed{R = 3} \]