# Thread: find the radius of convergence for 1st derivative of f.

1. ## find the radius of convergence for 1st derivative of f.

If f = n=0, infinite ((-1)^(n+1))(x-1)^(n))/n, find the radius of convergence for 1st derivative of f.

2. Originally Posted by ewkimchi
If f = n=0, infinite ((-1)^(n+1))(x-1)^(n))/n, find the radius of convergence for 1st derivative of f.
$\displaystyle f(x) = \sum_{n = 0}^{\infty}\frac{(-1)^{n + 1}(x - 1)^n}{n}$

$\displaystyle f'(x) = \sum_{n = 0}^{\infty}\frac{(-1)^{n + 1}n(x - 1)^{n - 1}}{n}$

$\displaystyle = \sum_{n = 0}^{\infty}(-1)^{n + 1}(x - 1)^{n - 1}$.

The radius of convergence of $\displaystyle f'(x)$ is given where

$\displaystyle \lim_{n \to \infty}\left|\frac{t_{n + 1}}{t_n}\right| < 1$

$\displaystyle \lim_{n \to \infty}\left|\frac{(-1)^{n + 2}(x - 1)^{n}}{(-1)^{n + 1}(x - 1)^{n - 1}}\right| < 1$

$\displaystyle \lim_{n \to \infty}\left|x - 1\right|< 1$

$\displaystyle |x - 1| < 1$

$\displaystyle -1 < x - 1 < 1$

$\displaystyle 0 < x < 2$.

The radius of convergence is $\displaystyle 1$, and the interval of convergence is $\displaystyle 0 < x < 2$.