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Question Number 198304 by universe last updated on 17/Oct/23

for {a_{n}} be a sequence of positive real numbers

such that a_{1} = 1, a_{n + 1}^{2} - {2 a}_{n} a_{n + 1} - a_{n} = 0, \forall n ⩾ 1

than the sum of series \underset{n = 1}{\sum^{\infty}} \frac{a_{n}}{3^{n}} lies in the interval

(A) (1, 2] (B) (2, 3] (C) (3, 4] (D) (4, 5]

Commented by Frix last updated on 17/Oct/23

a_{n + 1} = a_{n} + \sqrt{a_{n}^{2} + a_{n}} \Rightarrow a_{n} is strictly increasing

a_{n + 1} \sim 2 a_{n} + \frac{1}{2} for great n \land a_{n + 1} < 2 a_{n} + \frac{1}{2} \forall n \in N

b_{n + 1} = 2 b_{n} + \frac{1}{2} \land b_{1} = 1 \Rightarrow b_{n} = \frac{2^{n} 3 - 2}{4}

a_{n} < b_{n}

\sum_{n ⩾ 1} \frac{b_{n}}{3^{n}} = \lim_{n \to \infty} (\frac{5}{4} + \frac{1}{3^{n} 4} - \frac{2^{n - 1}}{3^{n - 1}}) = \frac{5}{4}

\Rightarrow

\sum_{n ⩾ 1} \frac{a_{n}}{3^{n}} < \frac{5}{4}

Commented by Frix last updated on 17/Oct/23

Same path with b_{2} = 1 + \sqrt{2} = a_{2} \Rightarrow \sum_{n ⩾ 1} \frac{a_{n}}{3^{n}} < \frac{3}{4} + \frac{\sqrt{2}}{3}

Commented by mr W last updated on 17/Oct/23

a_{n} = \frac{1}{2^{\frac{1}{2^{n - 1}}} - 1}

Commented by universe last updated on 17/Oct/23

s i r h o w t o f i n d t h i s r e s u l t ?

Commented by mr W last updated on 17/Oct/23

a_{n + 1}^{2} - 2 a_{n + 1} a_{n} - a_{n} = 0

{(\frac{1}{a_{n + 1}})}^{2} + \frac{2}{a_{n + 1}} - \frac{1}{a_{n}}

l e t b_{n} = \frac{1}{a_{n}}

b_{n + 1}^{2} + 2 b_{n + 1} - b_{n} = 0

b_{n + 1} = - 1 + \sqrt{1 + b_{n}}

b_{n + 1} + 1 = \sqrt{b_{n} + 1}

l e t c_{n} = b_{n} + 1 = \frac{1}{a_{n}} + 1

\Rightarrow c_{n + 1} = c_{n}^{\frac{1}{2}}

\Rightarrow c_{n} = {(c_{n - 1})}^{\frac{1}{2}} = {(c_{n - 2})}^{\frac{1}{2^{2}}} = \dots = c_{1}^{\frac{1}{2^{n - 1}}}

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

\Rightarrow c_{n} = 2^{\frac{1}{2^{n - 1}}} = \frac{1}{a_{n}} + 1

\Rightarrow a_{n} = \frac{1}{2^{\frac{1}{2^{n - 1}}} - 1}

Commented by Frix last updated on 18/Oct/23

Nice!

This can be written as

a_{n} = \underset{k = 0}{\sum^{2^{n - 1} - 1}} 2^{\frac{k}{2^{n - 1}}}

a_{1} = 2^{\frac{0}{1}} = 1

a_{2} = 2^{\frac{0}{2}} + 2^{\frac{1}{2}}

a_{3} = 2^{\frac{0}{4}} + 2^{\frac{1}{4}} + 2^{\frac{2}{4}} + 2^{\frac{3}{4}}

a_{4} = 2^{\frac{0}{8}} + 2^{\frac{1}{8}} + 2^{\frac{2}{8}} + 2^{\frac{3}{8}} + 2^{\frac{4}{8}} + 2^{\frac{5}{8}} + 2^{\frac{6}{8}} + 2^{\frac{7}{8}}

\dots

Commented by universe last updated on 18/Oct/23

t h a n k s s i r