lim-n-2-2-n-fractional-

Question Number 128626 by john_santu last updated on 09/Jan/21

\lim_{n \to \infty} {{(2 + \sqrt{2})}^{n}} = ?

{} = fractional

Commented by liberty last updated on 09/Jan/21

{x} = x - ⌊ x ⌋

Answered by mnjuly1970 last updated on 09/Jan/21

s o l u t i o n :

t_{n} := {(2 + \sqrt{2})}^{n} + {(2 - \sqrt{2})}^{n} \overset{w h y ?}{\in} N

b e c a u s e : t_{n} = \underset{k = 0}{\sum^{n}} {(\begin{matrix} n \\ k \end{matrix}) {(2)}^{n - k} {(\sqrt{2})}^{k} (1 + {(- 1)}^{k})

= \underset{k \underset{k \in N_{E}}{=} 0}{\sum^{n}} {(\begin{matrix} n \\ k \end{matrix}) 2^{n - k + \frac{k}{2} + 1}}

= \sum_{k \underset{k \in N_{E}}{=} 0} {(\begin{matrix} n \\ k \end{matrix}) 2^{[n + \frac{k}{2} + 1 \in N]}} \in N

b u t w e k n o w : 0 < {(2 - \sqrt{2})}^{n} < 1

∴ 0 < 1 - {(2 - \sqrt{2})}^{n} < 1

{l i m}_{n \to \infty} (1 - {(2 - \sqrt{2})}^{n}) = 1

{{(2 + \sqrt{2})}^{n}} = {t_{n} - {(2 - \sqrt{2})}^{n}}

= {t_{n} - 1 + 1 - {(2 - \sqrt{2})}^{n}}

∴ {l i m}_{n \to \infty} {{(2 + \sqrt{2})}^{n}} = 1

n o t e : {t_{n} - 1} = 0