Given-the-sequences-u-n-n-N-and-v-n-n-N-defined-by-u-n-k-0-n-1-k-and-v-n-u-n-1-n-n-a-Show-that-u-n-n-is-of-Cauchy-Deduce-that-u-n-n-converges-b-Show-that-u-n

Question Number 97781 by Ar Brandon last updated on 09/Jun/20

Given the sequences {(u_{n})}_{n \in N} and {(v_{n})}_{n \in N} defined

by u_{n} = \underset{k = 0}{\sum^{n}} \frac{1}{k!} and v_{n} = u_{n} + \frac{1}{n (n!)}

a ∖ Show that {(u_{n})}_{n} is of Cauchy . D educe that

{(u_{n})}_{n} converges .

b ∖ Show that {(u_{n})}_{n} and {(v_{n})}_{n} are adjacent

c ∖ Show that their common limit is not a rational

number .

Answered by maths mind last updated on 10/Jun/20

f o r \forall k ⩾ 2 k! ⩾ k (k - 1)

U_{n + m} - U_{n} = \underset{k = 1}{\sum^{m}} U_{k + n} ⩽ \underset{k = 1}{\sum^{m}} \frac{1}{(k + n) (k + n - 1)} = \underset{k = 1}{\sum^{m}} (\frac{1}{n + k - 1} - \frac{1}{n + k})

= \frac{1}{m} - \frac{1}{n + m} ⩽ \frac{1}{m} \dots . 1

U_{n} c a u c h y \Leftrightarrow \forall ϵ > 0 \exists N, \forall n, m ⩾ N ∣ U_{n} - U_{m} ∣< ϵ

w e c h o o s e N = E (\frac{1}{ϵ}) + 1 \Rightarrow U_{n} - U_{m} ⩽ \frac{1}{m} ⩽ \frac{1}{N} < ϵ \dots . b y 1

U_{n} i s c a u c h y s i n c e i s r e e l s e q u e n c e \Rightarrow U_{n} c o n v e r g e

b) U_{n + 1} - U_{n} = \frac{1}{(n + 1)!} > 0 U_{n} i s i b c r e a s i n g

V_{n + 1} - V_{n} = \frac{1}{(n + 1)!} + \frac{1}{(n + 1) (n + 1)!} - \frac{1}{n . n!}

\frac{n + 2 - (n + 1) (n + 1)}{(n + 1) (n + 1)!} = \frac{- n^{2} - n + 1}{(n + 1) (n + 1)!} < 0, \forall n ⩾ 1 V_{n} d e c r e a s i n g

U_{n} - V_{n} = \frac{1}{n n!} \to 0 U_{n}, V n a r e a d j a c e n t e

i f x = \frac{a}{b} = l i m U_{n} \in IQ

\underset{n = 0}{\sum^{+ \infty}} \frac{1}{n!} = \frac{a}{b} = \frac{a (b - 1)!}{b!} = \sum_{k ⩾ 0} \frac{1}{k!}

\Leftrightarrow a (b - 1)! = b! . (\underset{k = 0}{\sum^{b}} \frac{1}{k!}) + \frac{1}{b + 1} + \sum_{k ⩾ 2} \frac{b!}{(b + k)!}

k ⩾ 2 \Rightarrow \frac{b!}{(b + k)!} ⩽ \frac{1}{(b + k) (b + k - 1)}

\Rightarrow \sum_{k ⩾ 2} \frac{b!}{(b + k)!} ⩽ \sum_{k ⩾ 2} (\frac{1}{(b + k) (b + k - 1)}) = \frac{1}{b + 1}

\Rightarrow∣ a (b - 1)! - b! . \underset{k = 0}{\sum^{b}} \frac{1}{k!} ∣⩽ \frac{2}{b + 1} < 1 i f b ⩾ 2 \dots . E

b u t a (b - 1)! \in N a n d b! . \underset{k = 0}{\sum^{b}} \frac{1}{k!} \in N

\Rightarrow∣ a (b - 1)! - b! . \underset{k = 0}{\sum^{b}} \frac{1}{k!} ∣\in N \Rightarrow E a b s u r d

\Rightarrow x \notin Q

Commented by Ar Brandon last updated on 10/Jun/20

T hanks for your method,