I would start with an example..
Consider the power series $$\sum_{n=1}^\infty\frac{(n+4)(x-2)^n}{7^n(n^2+11)}$$
Determine the interval of convergence of this power series. If the interval is bounded, be sure to determine whether the series converges at the endpoints.
Apply Ratio Test
$$\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{\frac{(n+5)(x-2)^{n+1}}{7^{n+1}((n+1)^2+11)}}{\frac{(n+4)(x-2)^n}{7^n(n^2+11)}}\right|=\left|\frac{7^n(n^2+11)(n+5)(x-2)^{n+1}}{7^{n+1}((n+1)^2+11)(n+4)(x-2)^n}\right|=\left|\frac{(n^2+11)(n+5)(x-2)}{7(n^2+2n+12)(n+4)}\right|$$
And take the limit we have:
$$\lim_{n\to\infty}\left|\frac{(n^2+11)(n+5)(x-2)}{7(n^2+2n+12)(n+4)}\right|=\left|\frac{x-2}{7}\right|\lim_{n\to\infty}\left|\frac{(n^2+11)(n+5)}{(n^2+2n+12)(n+4)}\right|$$
Since $\forall x>0,\frac{(n^2+11)(n+5)}{(n^2+2n+12)(n+4)}>0$ that $$\left|\frac{x-2}{7}\right|\lim_{n\to\infty}\left|\frac{(n^2+11)(n+5)}{(n^2+2n+12)(n+4)}\right|= \left|\frac{x-2}{7}\right|\lim_{n\to\infty}\frac{(n^2+11)(n+5)}{(n^2+2n+12)(n+4)}$$ $$=\left|\frac{x-2}{7}\right|\lim_{n\to\infty}\frac{n^2(1+\frac{11}{n^2})n(1+\frac{5}{n})}{n^2(1+\frac{2}{n}+\frac{12}{n^2})n(1+\frac{4}{n})}$$ $$=\left|\frac{x-2}{7}\right|(1)=L$$
By ratio test $L<1\rightarrow \sum_{n=1}^\infty\frac{(n+4)(x-2)^n}{7^n(n^2+11)}$ converges $$\left|\frac{x-2}{7}\right|(1)<1$$ $$\left|\frac{x}{7}-\frac{2}{7}\right|<1$$ $$\frac{2}{7}-1<\frac{x}{7}<\frac{2}{7}+1$$ $$2-7<x<2+7$$ $$-5<x<9$$
Then we check the endpoints
When $x=5$ we have:
$$\sum_{n=1}^\infty\frac{(n+4)(-7)^n}{7^n(n^2+11)} =\sum_{n=1}^\infty\frac{(n+4)(-1)^n}{n^2+11}$$
Also $\forall n>0,\frac{(n+4)}{n^2+11}$ is decreasing
And $$\lim_{n\rightarrow\infty}\frac{(n+4)}{n^2+11}=\lim_{n\rightarrow\infty}\frac{(1+\frac{4}{n})}{n(1+\frac{11}{n^2})}=0$$
Therefore by Alternating series test we can conclude that the series converges.
When $x=9$
$$\sum_{n=1}^\infty\frac{(n+4)(7)^n}{7^n(n^2+11)}=\sum_{n=1}^\infty\frac{n+4}{n^2+11}$$
Recall that $\sum_{n=1}^\infty\frac{1}{n}$ diverges
And $$\frac{\frac{n}{n^2+11}}{\frac{1}{n}}=\frac{n^2}{n^2+11}=\frac{n^2}{n^2(1+\frac{11}{n^2})}=1$$
By Limit Comparison Test we have: $$\sum_{n=1}^\infty\frac{n}{n^2+11}\text{ is divergent}$$
Since $$\sum_{n=1}^\infty\frac{n}{n^2+11}\le\sum_{n=1}^\infty\frac{n+4}{n^2+11}$$
By Basic comparison Test that: $$\sum_{n=1}^\infty\frac{n+4}{n^2+11}\text{is divergent}$$
Therefore the interval of convergence is $-5\le x<9$
Finally I got the interval..normally, this is how I do this kind of questions, but sometimes in exam, the questions don't ask for steps.. so I just wondering if there is any quick way to determine interval of convergence for power series
What I guessed so far..
A power serie centered at $c$ defined as following: $$\sum _{n=0}^{\infty }\:a_n\left(x-c\right)^n$$ $1.$If $a_n=1$ have $$\sum _{n=0}^{\infty }\:\left(x-c\right)^n\text{ which converges on } (c-1,c+1)$$ $2.$If $a_n=\left(\frac{1}{a}\right)^n$ where $a\in\mathbb{R}\wedge a\neq0$ we have $$\sum _{n=0}^{\infty }\:\left(\frac{1}{a}\right)^n\left(x-c\right)^n\text{ which converges on } (c-a,c+a)$$ $3.$If $a_n=n^{b}\left(\frac{1}{a}\right)^n$ where $a,b\in\mathbb{R}\wedge a\neq0$ we have $$\sum _{n=0}^{\infty }\:n^{b}\left(\frac{1}{a}\right)^n\left(x-c\right)^n\text{ which converges on }\left\{\begin{array}{l} (c-a,c+a),a\in\mathbb{R}\wedge b\ge0 \\ [c-a,c+a),a>0\wedge -1\le b<0 \\ (c-a,c+a],a<0\wedge -1\le b<0 \\ [c-a,c+a],a\in\mathbb{R}\wedge b<-1\end{array}\right.$$ $4.$If $a_n=j^{n^k}n^b\left(\frac{1}{a}\right)^n$ where $a,b,j,k\in\mathbb{R}\wedge a\neq0$ we have $$\sum _{n=0}^{\infty }\:j^{n^k}n^b\left(\frac{1}{a}\right)^n\left(x-c\right)^n\text{ which converges on } \left\{\begin{array}{l}( \frac{cj-a}{j},\frac{cj+a}{j}),a\in\mathbb{R}\wedge b\ge0\wedge k=1 \wedge j\neq0 \\ [\frac{cj-a}{j},\frac{cj+a}{j}),a>0\wedge -1\le b<0\wedge k=1 \wedge j\neq0 \\ (\frac{cj-a}{j},\frac{cj+a}{j}],a<0\wedge -1\le b<0\wedge k=1 \wedge j\neq0 \\ [\frac{cj-a}{j},\frac{cj+a}{j}],a\in\mathbb{R}\wedge b<-1\wedge k=1 \wedge j\neq0 \\ [c,c],b\in\mathbb{N},a\in\mathbb{R}\wedge k>1 \wedge j\neq0 \\ \vdots \\ (-\infty,\infty),a\in\mathbb{R}\wedge b\in\mathbb{N},k\in\mathbb{R},j=0 \end{array}\right.$$
There could be more, but I didn't prove them yet and it's already getting too complicated for me to guess the formula, so I would stop here for now..
Reconsider the question $$\sum_{n=1}^\infty\frac{(n+4)(x-2)^n}{7^n(n^2+11)}= \sum_{n=1}^\infty\frac{(n+4)}{(n^2+11)}(\frac{1}{7})^n(x-2)^n$$ For first part only consider the power of $n$ $$\frac{(n+4)}{(n^2+11)}\text{ power of $n$ is $-1$}$$ It would have same interval of convergence as following $$\sum_{n=1}^\infty n^{-1}(\frac{1}{7})^n(x-2)^n$$
Use formula $3.$ Since $a=7,b=-1,c=2$, we can conclude that: $$\text{interval of convergence is }[c-a,c+a)=[-5,9)$$
But again, I didn't prove them yet, so there could be counter examples.. please tell me if there is any books talked about this.
Any help or suggestion would be appreciated.
Start by writing $y=x-2$ so that the series is $$\sum_{n=1}^\infty\frac{(n+4)y^n}{7^n(n^2+11)}$$ Then $${a_{n+1}\over a_n}=\frac17{n+5\over n+4}{{n^2+11}\over (n+1)^2+11}\to\frac17 \text{ as } n\to\infty$$ so the radius of convergence is $7$, and the interior of the interval is $(-5,9)$. There is no need to do all that calculation; surely you recognize by now that those fractions go to $1$, and I should think your teacher expects you to.
The series diverges at $x=9$ by the limit comparison test, but again you are doing a lot more than is needed. Just note that $$\lim_{n\to\infty} {{n+4\over n^2+11}\over \frac1n}=1$$ and you are done.
I wouldn't bother memorizing any formulas. Your basic conclusions are correct. Good job. The important thing is the coefficients, and when you have a rational function, what matters is the difference between the degree of the numerator and the degree of the denominator.