True or false.If the series converges for x=1.1, then it converges for x=7 I saw this question in a previous year test and it seemed pretty simple, and that can often mean that I am missing something.
If the series $$\sum_{n=0}^{\infty}a_n(x-3)^n$$ converges for $x=-1.1$, then it converges for $x=7$.
So, by definition, I can say that this series converges for $\lvert x-3\rvert<r$,. 
I can rewrite that as $-r+3<x<r+3$. 
Since r must be greater than $0$, I solve for $r$ in the left side, as the right side would yield me a negative $r$, so that $-r+3=-1.1$, which yields $r=4.1$. 
With this, I conclude that the series converges for $-1.1<r<7.1$ which makes the proposition true.
Is this procces right? Or am I missing something important?
 A: Your reasoning is not quite right because you did not solve the compound inequality correctly and you did not actually prove that $|x-3|<r$ for $x=7$.
Let's say we have the following, as you did:
$$-r+3 < x < r+3$$
Then, we need to solve this compound inequality for $x=-1.1$.
If $-r+3 < -1.1$, then $-r < -4.1$ and $r > 4.1$, or $4.1 < r$ If $-1.1 < r+3$, then $-4.1 < r$. Thus, these inequalities can be compounded to $-4.1 < 4.1 < r$ which can be simplified to $4.1 < r$.
Now, let's say $x=7$. If $x=7$, then $|x-3|=4$. Since $|x-3| < 4.1$ and $4.1 < r$, $|x-3|<r$, meaning the series converges for $x=7$, concluding the proof.
A: The related theorem is that if a power series 
$$
\sum a_k(z-z_0)^k
$$
converges for some $z=z_1$ then it also converges for all $z$ with
$$
|z-z_0|<|z_1-z_0|.
$$
Here $z_0=3$, $z_1=-1.1$ so that convergence is guaranteed for $|z-3|<4.1$. $z=7$ satisfies this inequality.

Proof idea: The terms of the $z_1$ series converge to zero, thus are bounded by some $M$ and thus 
$$
|a_k(z-z_0)^k|\le M·\left|\frac{z-z_0}{z_1-z_0}\right|^k.
$$
