a-0-k-a-n-1-1-1-a-n-lim-n-a-n-

Question Number 2297 by prakash jain last updated on 14/Nov/15

a_{0} = k

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

\lim_{n \to \infty} a_{n} = ?

Commented by Yozzi last updated on 15/Nov/15

a_{1} = \frac{1}{1 + a_{0}} = \frac{1}{1 + k} = \frac{(0) k + 1}{k + 1}

a_{2} = \frac{1}{1 + a_{1}} = \frac{1}{1 + \frac{1}{1 + k}} = \frac{k + 1}{k + 2}

a_{3} = \frac{1}{1 + a_{2}} = \frac{1}{1 + \frac{k + 1}{k + 2}} = \frac{k + 2}{2 k + 3}

a_{4} = \frac{1}{1 + a_{3}} = \frac{1}{1 + \frac{k + 2}{2 k + 3}} = \frac{2 k + 3}{3 k + 5}

a_{5} = \frac{1}{1 + a_{4}} = \frac{1}{1 + \frac{2 k + 3}{3 k + 5}} = \frac{3 k + 5}{5 k + 8}

a_{6} = \frac{1}{1 + a_{5}} = \frac{1}{1 + \frac{3 k + 5}{5 k + 8}} = \frac{5 k + 8}{8 k + 13}

a_{n} = \frac{F_{n - 1} k + F_{n}}{F_{n} k + F_{n + 1}}, n ⩾ 1

F_{i} = \frac{1}{\sqrt{5}} [{(\frac{1 + \sqrt{5}}{2})}^{i} - {(\frac{1 - \sqrt{5}}{2})}^{i}], i ⩾ 0

L e t r_{i} (+) = {(\frac{1 + \sqrt{5}}{2})}^{i}, r_{i} (-) = {(\frac{1 - \sqrt{5}}{2})}^{i}

∴ F_{i} = \frac{1}{\sqrt{5}} (r_{i} (+) - r_{i} (-))

∴ a_{n} = \frac{\frac{1}{\sqrt{5}} (k [r_{n - 1} (+) - r_{n - 1} (-)] + r_{n} (+) - r_{n} (-))}{\frac{1}{\sqrt{5}} (k [r_{n} (+) - r_{n} (-)] + r_{n + 1} (+) - r_{n + 1} (-))}

a_{n} = \frac{k (r_{n - 1} (+) - r_{n - 1} (-)) + r_{n} (+) - r_{n} (-)}{k (r_{n} (+) - r_{n} (-)) + r_{n + 1} (+) - r_{n + 1} (-)}

a_{n} = \frac{k (\frac{1}{r_{2} (+)} - \frac{r_{n - 1} (-)}{r_{n + 1} (+)}) + \frac{1}{r_{1} (+)} - \frac{r_{n} (-)}{r_{n + 1} (+)}}{k (\frac{1}{r_{1} (+)} - \frac{r_{n} (-)}{r_{n + 1} (+)}) + 1 - \frac{r_{n + 1} (-)}{r_{n + 1} (+)}}

\frac{r_{n} (-)}{r_{n + 1} (+)} = \frac{{(\frac{1 - \sqrt{5}}{2})}^{n}}{{(\frac{1 + \sqrt{5}}{2})}^{n + 1}} = {(\frac{1 - \sqrt{5}}{1 + \sqrt{5}})}^{n} \times \frac{2}{1 + \sqrt{5}}

∣ \frac{1 - \sqrt{5}}{1 + \sqrt{5}} ∣< 1 \Rightarrow {(\frac{1 - \sqrt{5}}{1 + \sqrt{5}})}^{n} \to 0 a s n \to \infty

\Rightarrow \lim_{n \to \infty} \frac{r_{n} (-)}{r_{n + 1} (+)} = 0

\frac{r_{n - 1} (-)}{r_{n + 1} (+)} = {(\frac{1 - \sqrt{5}}{1 + \sqrt{5}})}^{n} \times \frac{4}{(1 + \sqrt{5}) (1 - \sqrt{5})}

∴ \lim_{n \to \infty} \frac{r_{n - 1} (-)}{r_{n + 1} (+)} = 0

\frac{r_{n + 1} (-)}{r_{n + 1} (+)} = {(\frac{1 - \sqrt{5}}{1 + \sqrt{5}})}^{n + 1}

∴ \lim_{n \to \infty} \frac{r_{n + 1} (-)}{r_{n + 1} (+)} = 0

∴ \lim_{n \to \infty} a_{n} = \frac{\frac{k}{r_{2} (+)} + \frac{1}{r_{1} (+)}}{\frac{k}{r_{1} (+)} + 1} = \frac{{k r}_{1} (+) + r_{2} (+)}{r_{2} (+) (k + r_{1} (+))}

(C o m m e n t t h e m i s t a k e p l e a s e)

Commented by Yozzi last updated on 14/Nov/15

S o r r y . M i s t a k e .

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

If the value of k is such that a_{n} \neq - 1, \forall n \in N

then the limit is always \frac{\sqrt{5} - 1}{2}

Commented by Yozzi last updated on 14/Nov/15

I s e e . N i c e!

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

If the limits exists (when a_{n} \neq - 1 \forall n \in N)

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

a_{n} = \frac{1}{1 + a_{n}} \Rightarrow a_{n} = \frac{- 1 + \sqrt{5}}{2}