Consider the series ​

[infinity]
∑ (-1)^n (x+1)^n/ (n+1)!
n=0

a. Find the​ series' radius and interval of convergence. ​
b. For what values of x does the series converge​ absolutely?
c. For what values of x does the series converge​ coditionally?

Respuesta :

(a) If

[tex]f(x)=\displaystyle \sum_{n=0}^\infty (-1)^n \frac{(x+1)^n}{(n+1)!}[/tex]

then by the ratio test, the series converges for all x, since

[tex]\displaystyle \lim_{n\to\infty} \left| \frac{ (-1)^{n+1} \frac{(x+1)^{n+1}}{(n+2)!} }{ (-1)^n \frac{(x+1)^n}{(n+1)!} } \right|=\lim_{n\to\infty}\frac{|x+1|}{n+2}=0[/tex]

so the series radius of convergence is and the interval of convergence is (-∞, ∞).

(b) The series converges everywhere absolutely, because the ratio test for

[tex]\displaystyle \sum_{n=0}^\infty \frac{|x+1|^n}{(n+1)!}[/tex]

also shows the radius of convergence is ∞.

(c) The series converges absolutely, so conditional convergence is moot.