Does-1-i-1-1-i-converge-2-i-1-1-i-converge-

Question Number 5921 by FilupSmith last updated on 05/Jun/16

Does :

(1) \underset{i = 1}{\sum^{\infty}} Γ (\frac{1}{i}) converge ?

(2) \underset{i = 1}{\prod^{\infty}} Γ (\frac{1}{i}) converge ?

Commented by Yozzii last updated on 05/Jun/16

(1) Γ (x) Γ (1 - x) = \frac{π}{s i n π x}

Γ (i^{- 1}) = \frac{π}{Γ (1 - i^{- 1}) s i n \frac{π}{i}}

Γ (x) = \int_{0}^{\infty} u^{x - 1} e^{- t} d t \Rightarrow Γ (1) = \int_{0}^{\infty} e^{- t} d t = 1

\lim_{i \to \infty} Γ (i^{- 1}) = \frac{π}{Γ (1) s i n π \times 0} = \frac{π}{Γ (1) \times 0} = \frac{π}{1 \times 0} = \frac{π}{0} \neq 0

S i n c e \lim_{i \to \infty} Γ (i^{- 1}) \neq 0, \underset{i = 1}{\sum^{\infty}} Γ (i^{- 1}) i s n o t c o n v e r g e n t .

(i i) L e t (a_{n}) b e a s e q u e n c e w i t h e a c h

a_{n} > 0 . I f \underset{n = 1}{\prod^{\infty}} a_{n} c o n v e r g e s t o a l i m i t

l a n d l \neq 0, t h e n (a_{n}) c o n v e r g e s a n d

\lim_{n \to \infty} a_{n} = 1 .

\Rightarrow I f (a_{n}) d i v e r g e s, s o t h a t \lim_{n \to \infty} a_{n} \neq 1,

t h e n \underset{n = 1}{\prod^{\infty}} a_{n} d i v e r g e s a l s o .

S i n c e \lim_{i \to \infty} Γ (i^{- 1}) d o e s n o t e x i s t a n d Γ (i^{- 1}) > 0 f o r i \in N,

t h e n \underset{i = 1}{\prod^{\infty}} Γ (i^{- 1}) d i v e r g e s .

Commented by FilupSmith last updated on 05/Jun/16

Amazing!