Interval of Convergence

[ginnyflute453]

Let f(x) be a function defined by f(x)= (sum from n=1 to infinity) ((-1)[sup:349j3za1]n+1[/sup:349j3za1]*x[sup:349j3za1]n[/sup:349j3za1])/(4[sup:349j3za1]n[/sup:349j3za1]))

a. Determine the interval of convergence for f(x)
b. Determine the interval of convergence for (integral) of f(x)dx

 

[BigGlenntheHeavy]

$\displaystyle f(x) \ = \ \sum_{n=1}^{\infty}\frac{(-1)^{n+1} x^{n}}{4^{n}}$

$\displaystyle Ratio \ Test: \ \lim_{n\to\infty}\bigg|\frac{a_{n+1}}{a_n}\bigg| \ = \ \lim_{n\to\infty}\bigg|\frac{x^{n+1}}{4^{n+1}}*\frac{4^{n}}{x^{n}}\bigg|$

$\displaystyle = \ \lim_{n\to\infty}\frac{|x|}{4} \ < \ 1, \ \implies \ |x| \ < \ 4 \ \implies \ -4 \ < \ x \ < \ 4.$

$\displaystyle Hence, \ interval \ of \ convergence \ is \ -4 \ < \ x \ < \ 4, \ but \ what \ about \ the \ endpoints?$

$\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n+1}(-4)^{n}}{4^{n}} \ = \ \sum_{n=1}^{\infty}\frac{(-1)^{n+1}(-1)^{n}4^{n}}{4^{n}}$

$\displaystyle = \ \sum_{n=1}^{\infty}(-1)^{2n+1} \ = \ -1-1-1-1-1... \ diverges, \ and$

$\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n+1}4^{n}}{4^{n}} \ = \ \sum_{n=1}^{\infty}(-1)^{n+1} \ = \ 1-1+1-1+1...diverges$

$\displaystyle Now, \ since \ both \ endpoints \ diverge, \ the \ limit \ of \ convergence \ is \ (-4,4).$