find-radius-andsum-of-n-1-n-1-n-x-n-

Question Number 29983 by abdo imad last updated on 14/Feb/18

f i n d r a d i u s a n d s u m o f \sum_{n = 1}^{\infty} \frac{n - 1}{n!} x^{n} .

Commented by abdo imad last updated on 15/Feb/18

S (x) = \sum_{n = 1}^{\infty} \frac{n - 1}{n!} x^{n} = \sum_{n = 1}^{\infty} u_{n} (x) a n d f o r x \neq 0

∣ \frac{u_{n + 1} (x)}{u_{n} (x)} ∣= ∣ \frac{\frac{n}{(n + 1)!} x^{n + 1}}{\frac{n - 1}{n!} x^{n}} ∣= \frac{n}{(n + 1)!} . \frac{n!}{n - 1} ∣ x ∣= \frac{n}{(n + 1) (n - 1)} ∣ x_{} ∣_{n \to + \infty} \to 0

s o R = + \infty w e h a v e S (x) = \sum_{n = 1}^{\infty} \frac{x^{n}}{(n - 1)!} - \sum_{n = 1}^{\infty} \frac{x^{n}}{n!} b u t

\sum_{n = 1}^{\infty} \frac{x^{n}}{n!} = e^{x} - 1 a n d \sum_{n = 1}^{\infty} \frac{x^{n}}{(n - 1)!} = x \sum_{n = 1}^{\infty} \frac{x^{n - 1}}{(n - 1)!}

= x \sum_{n = 0}^{\infty} \frac{x^{n}}{n!} = {x e}^{x} \Rightarrow S (x) = x e^{x} - e^{x} + 1 = (x - 1) e^{x} + 1 .