Approaching Maclaurin/Taylor series from a different angle. So I have been asked a question for one of my problem sets but I haven't found a "good" way to approach it. Personally I was thinking of approaching the series it backwards but then again if I want to use a certain test(i.e. ratio test, comparison test, square root test, etc etc) that may be a much more difficult. Here it is,
For each interval given below, build a power series which has that interval as its interval of convergence. Justify your claims by showing that the interval of convergence of your series is as desired.
(a) [0,4)
(b) [-13,-6]
(c) (11,12)
If anybody can give me a good suggestion on how to approach the problem that would be great!
 A: Example (a): The midpoint of the interval is $2$, so we will be looking at something of shape $\sum a_n(x-2)^n$. We want radius of convergence $2$, so let us look for something of shape $\sum b_n \frac{(x-2)^n}{2^n}$. 
We want convergence at $0$, and divergence at $4$.   Look for something like $b_n=\frac{1}{n}$, possibly decorated with $(-1)^n$. 
Let's check out $\sum \frac{1}{n2^n}(x-2)^n$. Now we can work in the standard direction, as in the usual find the interval of convergence question.  Apply the Ratio Test. Then see what happens at $x=0$. Alternating series, convergence. At $x=4$, things go bad, we get the harmonic series.
Example (b): Use the same basic strategy, but use $b_n=\frac{1}{n^2}$ to ensure convergence at both ends. 
The last one is left to you. 
A: It is easy to translate and scale the interval of convergence of a power series, so all you really need to worry about is finding power series with the right behavior at the endpoints. For example, $\sum\limits_{n=0}^\infty x^n$ has interval of convergence $(-1,1)$, $\sum\limits_{n=0}^\infty \frac{x^n}{n}$ has interval of convergence $[-1,1)$ and $\sum\limits_{n=0}\frac{x^n}{n^2}$ has interval of convergence $[-1,1]$. All you need to do now is translate and scale. For a), we translate the interval of convergence of $\sum\limits_{n=0}^\infty \frac{x^n}{n}$ to the right by $1$ by substituting $x-1$ for $x$, and then scale this by $2$ by substituting $\frac{x}{2}$ for $x$, giving us $\sum\limits_{n=0}^\infty \frac{(x/2-1)^n}{n}$. The others can be solved similarly.
