Is this the correct convergence interval and radius of power series? I was requested to find convergence interval and radius of 
$$\sum_{n=1}^n \frac{e^n}{n^3}(4-x)^n$$
First of all, see that
$$a_n=\frac{e^n}{n^3}(4-x)^n=\frac{(e(4-x))^n}{n^3}=\frac{(e*-(x-4))^n}{n^3}=\frac{(-e)^n}{n^3}(x-4)^n$$
so that
$$\sum_{n=1}^n \frac{e^n}{n^3}(4-x)^n=\sum_{n=1}^n \frac{(-e)^n}{n^3}(x-4)^n$$
We therefore have a power series centered at $4$.
$a)$ Take $$|\frac{a_{n+1}}{a_n}|=|\frac{(-e)^{n+1}}{(n+1)^n}(x-4)^{n+1}*\frac{n^3}{(-e)^n(x-4)^n}|=|e(x-4)(\frac{n}{n+1})^3|$$
so that $\lim_{n\to\infty} |\frac{a_{n+1}}{a_n}|=e|x-4|$
$b)$ Because $e|x-4|<1$ we have that the series converges for $|x-4|<\frac{1}{e}$, and its radius is $R=\frac{1}{e}$. I will skip this to keep this short, but it is not hard to show that $|x-4|<\frac{1}{e} \iff x\in (4-\frac{1}{e}, 4+\frac{1}{e})$. (The fact that the serie converges for $x=c=4$ is important.)
If we examine the boundaries of the interval, we'll see that for $x=4-\frac{1}{e}$, we have a $p$ series with $p>1$ and it converges. 
For $x=4+\frac{1}{e}$ we have an alternating series with $b_n=\frac{1}{n^3}$, where $b_n$ is decreasing and $\lim_{n\to\infty} b_n = 0$. Therefore, according to the alternating series test, it converges. 
(I don't show this because it's as trivial as replacing $x$ with the values and see what we get; don't want to make this too long for anyone to read.)
Therefore the interval of convergence is $[4-\frac{1}{e}, 4+\frac{1}{e}]$ and $R=\frac{1}{e}$.

Yes, the radius of convergence is $\frac1e$ and the interval of convergence is $\left[4-\frac1e,4+\frac1e\right]$. You can prove that your series converges at both points $4\pm\frac1e$ by noting that if $x$ is any of these points, then$$\sum_{n=0}^\infty\left\lvert\frac{e^n}{n^3}(4-x)^n\right\rvert=\sum_{n=0}^\infty\frac1{n^3},$$which converges. So, your series converges absolutely when $x=4\pm\frac1e$.
