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Question Number 2286 by Filup last updated on 14/Nov/15

Can you evaluate :

\frac{1}{1} + \frac{1}{1 + \frac{1}{1}} + \frac{1}{1 + \frac{1}{1 + \frac{1}{1}}} + \dots + \frac{1}{1 + \frac{1}{\dots}}

Commented by Yozzi last updated on 14/Nov/15

u_{1} = \frac{1}{1}, u_{2} = \frac{1}{1 + \frac{1}{1}} = \frac{1}{1 + u_{1}}

u_{3} = \frac{1}{1 + \frac{1}{1 + \frac{1}{1}}} = \frac{1}{1 + u_{2}}

u_{4} = \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1}}}} = \frac{1}{1 + u_{3}}

R e c u r r e n c e e q u a t i o n f o r t e r m s i s

u_{n + 1} = \frac{1}{1 + u_{n}}, u_{1} = 1, n ⩾ 1

u_{1} = 1

u_{2} = \frac{1}{1 + u_{1}} = \frac{1}{2}

u_{3} = \frac{1}{1 + u_{2}} = \frac{1}{1 + \frac{1}{2}} = \frac{2}{3}

u_{4} = \frac{1}{1 + u_{3}} = \frac{1}{1 + \frac{2}{3}} = \frac{3}{5}

u_{5} = \frac{1}{1 + u_{4}} = \frac{1}{1 + \frac{3}{5}} = \frac{5}{8}

u_{6} = \frac{1}{1 + u_{5}} = \frac{1}{1 + \frac{5}{8}} = \frac{8}{13}

Commented by Rasheed Soomro last updated on 17/Nov/15

G^{OO} D APPROACH!

Commented by prakash jain last updated on 14/Nov/15

a_{n + 1} = \frac{1}{1 + a_{n}}

\lim_{n \to \infty} a_{n} = \frac{- 1 + \sqrt{5}}{2}

Sum {doesn}^{'} t converge .

Answered by 123456 last updated on 15/Nov/15

by the coments

u_{n + 1} = \frac{1}{1 + u_{n}}, u_{0} = 1

claim 1 :

u_{n} = \frac{F_{n}}{F_{n + 1}} F_{n} are Fibonnaci numbers

proof :

F_{n + 2} = F_{n + 1} + F_{n}

for n = 1

u_{1} = \frac{F_{1}}{F_{2}} = \frac{1}{1} = 1 (T)

suppose its truth for n, lets proof for n +

u_{n + 1} = \frac{1}{1 + u_{n}} = \frac{1}{1 + \frac{F_{n}}{F_{n + 1}}}

= \frac{1}{\frac{F_{n} + F_{n + 1}}{F_{n + 1}}} = \frac{F_{n + 1}}{F_{n + 2}}

- - - - - - - - - - - - -

S = Σ u_{n}

a necesary condition to Σ u_{n} exist is

u_{n} \to 0 as n \to \infty (but not sulficient, 1 / n \to 0 but Σ 1 / n diverge)

but the sum seem to be divergent

\frac{F_{n + 1}}{F_{n}} \to φ = \frac{1 + \sqrt{5}}{2}

u_{n} = \frac{F_{n}}{F_{n + 1}} = \frac{1}{\frac{F_{n + 1}}{F_{n}}} \to \frac{1}{φ} = \frac{\sqrt{5} - 1}{2}

Commented by prakash jain last updated on 15/Nov/15

\lim_{n \to \infty} u_{n} = \frac{- 1 + \sqrt{5}}{2} o r \frac{1 + \sqrt{5}}{2} ?

Commented by 123456 last updated on 15/Nov/15

\frac{2}{1 + \sqrt{5}} = \frac{2 (1 - \sqrt{5})}{(1 + \sqrt{5}) (1 - \sqrt{5})} = \frac{2 (1 - \sqrt{5})}{1 - 5} =

\frac{2 (1 - \sqrt{5})}{- 4} = \frac{\sqrt{5} - 1}{2}

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