TITLE: Radius of convergence of power series or geometric series
QUESTION [5 upvotes]: To find radius of convergence of geometric series $$\sum_{n=1}^\infty a_n$$ I need to use ratio/root test to find $|L|&lt1$
To find radius of convergence of power series $$\sum_{n=1}^\infty c_n (x-a)^n$$ I am supposed to find the limit $L$ of just the constant term $c_n$?
$$\sum_{n=1}^\infty \frac{(2x-5)^n}{n^2}, \qquad c_n = \frac{2^n}{n^2}, \qquad R = 2^{-1}=1/2$$
How do I know what is a constant? Free of $n$? Or I should focus on getting it into the form $\sum_{n=1}^\infty c_n (x-a)^n$. Thats what I did below
$$\sum_{n=1}^\infty \frac{(x-1)^{n-3} + (x-1)^{n-1}}{4^n + 2^{2n-1}}$$
I first factorized what I can
$$ = \sum_{n=1}^\infty (x-1)^n \cdot \frac{(x-1)^{-3} + (x-1)^{-1}}{4^n(1 + 2^{-1})}$$
So I guess $$c_n=\frac{(x-1)^{-3} + (x-1)^{-1}}{4^n(1 + 2^{-1})}$$
Then $$L=|\frac{c_{n+1}}{c_n}|$$ But what do I do with the $x$? 

Correct method is supposed to be knowing its a geometric series of common ratio $\frac{x-1}{4}$. Then radius is |x-1|<4. But here, don't I need to get the radius of $x$ alone, without the $-1$?
Anyways, I think the main thing I am confused about is when to use which method. For geometric series, I am doing the ratio/root test for the whole $a_n$ while in power series, I am "separating" the $a_n$ into $c_n (x-a)^n$ form? And doing test on the constant terms ($c_n$)?

REPLY [2 votes]: Without using ratio test u can also find radius of convergence in this example,
$\sum_{n=1}^\infty c_n$ $\frac{(2x-5)^n} {n^2}$
Comparing the given series with power series,
$\sum_{n=1}^{\infty}c_n(x-x_0)^n$
We have, $c_n=\frac{1}{2^n n^2}$ and $x_0=(\frac{5}{2})^n$
$R=\lim_{n\to \infty}|c_n|^\frac{1}{n}$
$\ \ \ =\lim_{n\to \infty}|\frac{1}{2[(n)^\frac{1}{n}]^2}| $
$\ \ \ R=\frac{1}{2}$
$\ \ \ R=\frac{1}{r}$
$\ \ \ r=2 $
$\therefore$ radius of convergence is 2
Moreover,interval of convergence $(x_0-r,x_0+r)$
In this example root test is very useful because here constant having power n