To find radius of convergence of geometric series $\sum_{n=1}^\infty a_n$ I need to use ratio/root test to find |L|<1
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 ($c_n$)?