Inferring interval of convergence from one converging value (x) and given center 
If $\sum_{n=0}^{\infty} c_n(x-5)^n$ converges when $x=3$, what can you say about the convergence of the following series?
(1) $\sum_{n=0}^{\infty} c_n(-1)^n$
(2) $\sum_{n=0}^{\infty} c_n2^n$

The answer states that (1) is convergent that nothing can be determined about (2).
I know I could apply the ratio test if I was asked to obtain the interval of convergence and a series such as $\sum \frac{x^2}{n2^n}$. The limit will need to be less than 1 for the series to converge. We'll get the inequality $\frac{|x|}{2} < 1$ which yields us $R = 2$ and the inequality $-2 < x < 2$.
I know that once we have an inequality like that, an $x$ value which holds the inequality would be convergent, and $x$ value that doesn't hold  would be divergent, and that the endpoints are inconclusive without further testing.
I don't understand how one can infer an interval of convergence from this problem.
A power series is defined as $\sum c_n(x - a)^n$ where the series centers around $a$. The radius can't be $0$ because we have an $x$ where $x \neq a$ ($3 \neq 5$). How do I know the series doesn't converge for all $x$, giving an interval of $(-\infty, \infty)$?
How do I infer a radius from knowing that $x = 3$ converges and that $a = 5$?
I figure that once I know a radius, I will have an inequality of $5 - R < x < 5 + R$.
 A: We're given that convergence holds at $x=3$, so from the definition of the radius of convergence, $R \ge 2$. 
For part (1), we have $-1 = 4-5$ and $|4-5| = 1 < 2$, so it converges.
Now, for part (2) we need to be a little more careful. Since $R \ge 2$, we can either have $R > 2$, in which case we note $2 = 7-5$ and $|7-5| = 2 < R$ and the series converges, or we can say that $R = 2$ and then the point $x = 7$ is on the boundary. If the point is on the boundary, we cannot conclusively determine convergence behavior, and it is true that different points on the boundary can behave differently (e.g. converge at $x=3$ and diverge at $x = 7$).
Since convergence is guaranteed in the case $R > 2$ but not guaranteed in the case $R = 2$, we cannot make a conclusion.
A: We know that $x = 3$ is convergent so $R$ must be at least 2. However, we cannot determine what the radius is equal to.
\begin{align}
|x - a| &< R\\
-R + a < x &< R + a\\
-R + 5 < x &< R + 5&a&=5\\
3 \geq x &< 7&R &= 2
\end{align}
Now, the series converges for $|x - a| > R$ and is inconclusive for $|x - a| = R$. However, we're told that the series does converge at $x = 3$. This solves the inconclusion on the left-hand side of the last line, thus why it is $3 \geq x$ and not $3 < x$.
$$
\sum c_n(-1)^n \Rightarrow (x - 5) = -1 \Rightarrow x = 4\\
3 \geq x = 4 < 7
$$
This inequality holds, so $\sum c_n (-1)^n$ is convergent.
$$
\sum c_n2^n \Rightarrow (x - 5) = 2 \Rightarrow x = 7\\
3 \geq 7 < 7
$$
This inequality does not hold, and since $x$ is equal to our right endpoint, no conclusions can be made about the sum $\sum c_n 2^n$ .
$$\\\\$$
Another example for posterity, to hit the point home.

If $\sum_{n=0}^{\infty} c_n2^n$ is convergent, what can you conclude about the convergence of the following series?
(1) $\sum_{n=0}^{\infty} c_n(-2)^n$
(2) $\sum_{n=0}^{\infty} c_n(-3)^n$

If $x=2$ then $R$ must be at least 2.
\begin{align}
-R + a < x &< R + a\\
-R < x &< R&a&=0\\
-2 < x &\leq 2&R &= 2
\end{align}
Note again, the right-hand side is $x \leq 2$ instead of $x < 2$ for the same reason as before.
\begin{align}
-2 < x = -2 \leq 2&\quad(1)\\
-2 < x = -3 \leq 2&\quad(2)
\end{align}
Both of these are inconclusive. If the radius was greater than 2—and we have nothing to tell us it isn't—then $x = 2$ would still hold (as it would just be inside the interval of convergence), but we can't make any conclusions about $x=-2$ or $x=-3$.
