Given a power series with interval of convergence $(-1,1]$, construct a series with another given interval of convergence Suppose that you have a power series 
$$\sum_{n=1}^\infty (a_nx^n)$$
 whose interval of convergence is $(-1,1]$.  
A) Using the same numbers $(a_n)$, come up with a new power series whose interval of convergence is $(0,2]$ 
B) Using the same numbers $(a_n)$, come up with a new power series whose interval of convergence is $(-2,2]$. 
C) Using the same numbers $(a_n)$, come up with a new power series whose interval of convergence is $[-1,1)$. Hint: what transformation would turn the interval $(-1,1]$ into $[-1,1)$? 
D) Putting together ideas from the previous parts of this question, come up with a new power series whose interval of convergence is $[10,20)$ 
For part A I got 
$$\sum_{n=1}^\infty a_n(x-1)^n. $$
For part B I got 
$$\sum_{n=1}^\infty a_n(2x)^n.$$ 
Part C and D I have no idea. I'm really confused. Can someone explain? 
 A: Part b is not quite right. If you have a power series with convergence in $x$ on $(-1,1]$ then you should expand the region by shrinking the variable. In other words if $x=\frac{1}{2}y$ then the interval $(-1,1]$ in terms of $x$ corresponds to the interval $(-2,2]$ in terms of $y$. (To confirm, just consider the end points $x=-1$ and $x=1$).
For c, we know that the power series converges on $(-1,1]$. If we let $x=-y$ then at $x=1$ we have $y=-1$ and the range of $y$ is $[-1,1)$, so
$$\sum_{n=1}^{\infty} a_n x^n = \sum_{n=1}^{\infty} a_n (-y)^n$$
has a interval of convergence of $[-1,1)$ in terms of $y$.
In order to answer d you have to consider the steps taken in a, b, and c. We first 'expand' the interval from $(-1,1]$ in terms of $x$ to $(-15,15]$ in terms of $y$:
$$\sum_{n=1}^{\infty} a_n x^n = \sum_{n=1}^{\infty} a_n \left(\frac{y}{15}\right)^n$$
which may then be 'reflected around the center of the interval' to have an interval of convergence of $[-15,15)$ in terms of $z$:
$$\sum_{n=1}^{\infty} a_n \left(\frac{y}{15}\right)^n 
= \sum_{n=1}^{\infty} a_n \left(\frac{(-z)}{15}\right)^n$$
which we then 'shift' to $[-10,20)$ in terms of $w$:
$$\sum_{n=1}^{\infty} a_n \left(-\frac{z}{15}\right)^n = \sum_{n=1}^{\infty} a_n \left(-\frac{(w-5)}{15}\right)^n$$
or more simply
$$\sum_{n=1}^{\infty} a_n \left(\frac{5-w}{15}\right)^n< \infty \quad\text{for $w\in[-10,20)$}$$
Note, for a sanity check we may check the end points. $w=-10$ corresponds to $x=\frac{5-(-10)}{15} = 1$, where the series converges in terms of $x$. Also $w=20$ corresponds to $x=\frac{5-(20)}{15}=-1$ where we know that the series diverges. 
A: For part (c), consider the series:
$$\sum_{n=1}^\infty (a_nx^n(-1)^n)$$
Since when $x =1$, our series becomes: $\sum a_n$, therefore, the series $a_n$ converges.
Therefore, if we want $a_n$ to converge when $x= -1$, by adding $(-1)^n$, when $x =-1$, the series then becomes $\sum (a_n)(-1)^n$. Therefore, if it is multiplied by $(-1)^n$ it will cancel out with the other one and leave our answer to be: $\sum a_n$
You can use this idea to help you answer part (d). of this question, as they are similar in concept.
