Question

Consider the series ​

Answer 1

(a) If

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

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

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

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

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

$\displaystyle \sum_{n=0}^\infty \frac{|x+1|^n}{(n+1)!}$

also shows the radius of convergence is ∞.

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