need some help with a power series convergence test problem Find the interval of convergence for the given power series: 
$$\sum\limits_{n=1}^\infty \frac{(x - 1)^n }{n(-4)^n}$$
First I applied the generalized ratio test, came out with $\frac{(1-x)}{4}$
Solved the inequality $|1-x| \lt 4$ and got $-3 \lt x \lt 5$.
But webwork refuses to accept my answer.  Am I doing something wrong?

$$\sum\limits_{n=1}^{\infty} \frac{(x - 1)^n }{n(-4)^n}$$
Using the ratio test, we have
$$ \lim\limits_{n\to\infty} \left|\frac{\frac{(x - 1)^{n+1}}{(n+1)(-4)^{n+1}}}{\frac{(x - 1)^n }{n(-4)^n}}\right|$$
$$ =\lim\limits_{n\to\infty} \left|\frac{(x - 1)^{n+1}n(-4)^n}{(x-1)^n(n+1)(-4)^{n+1}}\right|$$
$$ =\lim\limits_{n\to\infty} \left|\frac{(x - 1)n}{(n+1)(-4)}\right|$$
$$ =\frac14\left|x-1\right|\lim\limits_{n\to\infty} \left|\frac{n}{n+1}\right|$$
$$ =\frac14\left|x-1\right|\lim\limits_{n\to\infty} \left|\frac{1}{1+\frac1n}\right|$$
$$ =\frac14\left|x-1\right|$$
Now let's find the interval of convergence 
$$ \frac14\left|x-1\right|\lt 1 $$
$$ \left|x-1\right|\lt 4 $$
$$ -4\lt x-1\lt 4 $$
$$ -3\lt x\lt 5 $$
At $x=-3$, we have
$$\sum\limits_{n=1}^{\infty} \frac{(-4)^n }{n(-4)^n}= \sum\limits_{n=1}^{\infty} \frac{1}{n}\Rightarrow \mbox{diverges} $$
At $x=5$, we have
$$\sum\limits_{n=1}^{\infty} \frac{4^n }{n(-4)^n}= \sum\limits_{n=1}^{\infty} \frac{(-1)^n}{n}\Rightarrow \mbox{converges} $$
Therefore, the interval of convergence is 
$$ -3\lt x\leq 5 $$
