convergence-radius-of-n-N-2-n-z-n-

Question Number 99496 by ~blr237~ last updated on 21/Jun/20

c o n v e r g e n c e r a d i u s o f \sum_{n \in N} 2^{n} z^{n!}

Answered by mathmax by abdo last updated on 21/Jun/20

u_{n} = 2^{n} z^{n!} is with positive terms

\frac{u_{n + 1}}{u_{n}} = \frac{2^{n + 1} z^{(n + 1)!}}{2^{n} z^{n!}} = 2 z^{(n + 1)! - n!} = 2 z^{(n + 1) n! - n!} = 2 z^{n (n!)} = 2 z^{φ (n)} (φ (n) \to \infty)

if ∣ z ∣< \frac{1}{2} \Rightarrow∣ {2 z}^{φ (n)} ∣< 1 and \lim_{n \to + \infty} \frac{u_{n + 1}}{u_{n}} = 0 \Rightarrow Σ u_{n} converges

anther {wayu}_{n}^{\frac{1}{n}} = 2 {(z^{n!})}^{\frac{1}{n}} = 2 {(z)}^{(n - 1)!} \Rightarrow if ∣ z ∣< \frac{1}{2} \Rightarrow \lim_{n \infty} u_{n}^{\frac{1}{n}} = 0 \Rightarrow Σ u_{n} converges

\Rightarrow r = \frac{1}{2}

Commented by maths mind last updated on 21/Jun/20

s i r i t h i n k R = 1

Σ 2^{n} {(\frac{2}{3})}^{n!}

\exists a < 1 & \exists N \in N ∣ \forall n ⩾ N

2^{n} {(\frac{2}{3})}^{n!} < a^{n}

\Leftrightarrow n! l n (\frac{2}{3}) < n l n (\frac{a}{2})

\Leftrightarrow (n - 1)! ⩾ \frac{l n (\frac{a}{2})}{l n (\frac{2}{3})}, a = \frac{1}{4}

(n - 1)! ⩾ \frac{l n (8)}{l n (\frac{3}{2})} h a s e s o l u t i o n s i n c e (n - 1)! \to \infty

Σ 2^{n} {(\frac{2}{3})}^{n!} = \underset{n = 0}{\sum^{N - 1}} 2^{n} {(\frac{2}{3})}^{n!} + \sum_{k ⩾ N} 2^{k} {(\frac{2}{3})}^{k}

f i r s t f i n i t s u m 2 n d

2^{k} {(\frac{2}{3})}^{k} < {(\frac{1}{4})}^{n} b y c o m p a r a i s o n C v

l e t a < 1

\exists N_{a} \in N ∣ \forall n ⩾ N 2^{n} {(a a a a)}^{n!} < {(\frac{a}{2})}^{n}

\Leftrightarrow n! l n (a) < n l n (\frac{a}{4}) \Leftrightarrow (n - 1)! > \frac{l n (\frac{a}{4})}{l n (a)} h a s e a l w a y s s o l u t i o n

f o r a \in] 0, 1 [

Commented by mathmax by abdo last updated on 22/Jun/20

perhaps sir i am not sure for my answer because this serie is more

complicated with this factoriel . .!