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Question Number 177741 by infinityaction last updated on 08/Oct/22

	Answered by Frix last updated on 08/Oct/22

	$let p = a_{n} \land q = a_{n + 1}$ $q^{2} - 2 p q - p = 0 \Rightarrow q = p + \sqrt{p (p + 1)}$ $\Rightarrow q > p \Leftrightarrow a_{n + 1} > a_{n}$ $\lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = \lim_{p \to \infty} \frac{p + \sqrt{p^{2} + p}}{p} = 2$ $\Rightarrow$ $\frac{a_{n}}{3^{n}} \sim \frac{2^{n}}{3^{n}}$ $\underset{n = 1}{\sum^{\infty}} \frac{2^{n}}{3^{n}} = 2$ $\Rightarrow$ $1 < \underset{n = 1}{\sum^{\infty}} \frac{a_{n}}{3^{n}} < 2$
	Commented by infinityaction last updated on 08/Oct/22

	$t h a n k s s i r$
	Answered by mr W last updated on 08/Oct/22

	$a_{n + 1}^{2} - 2 a_{n + 1} a_{n} - a_{n} = 0$ ${(\frac{a_{n + 1}}{a_{n}})}^{2} - 2 (\frac{a_{n + 1}}{a_{n}}) + 1 = \frac{1}{a_{n}} + 1$ ${(\frac{a_{n + 1}}{a_{n}} - 1)}^{2} = \frac{1}{a_{n}} + 1 > 1, s i n c e a_{n} > 0 .$ $\Rightarrow \frac{a_{n + 1}}{a_{n}} > 2$ ${(\frac{a_{n + 1}}{a_{n}} - 1)}^{2} = \frac{1}{a_{n}} + 1 < 2, s i n c e a_{n} > 1 .$ $\Rightarrow \frac{a_{n + 1}}{a_{n}} < 1 + \sqrt{2}$ $2 < \frac{a_{n + 1}}{a_{n}} < 1 + \sqrt{2}$ $2 < \frac{a_{n}}{a_{n - 1}} < 1 + \sqrt{2}$ $2 < \frac{a_{n - 1}}{a_{n - 2}} < 1 + \sqrt{2}$ $. . .$ $2 < \frac{a_{2}}{a_{1}} < 1 + \sqrt{2}$ $Π :$ $2^{n - 1} < \frac{a_{n}}{a_{1}} < {(1 + \sqrt{2})}^{n - 1}$ $\Rightarrow 2^{n - 1} < a_{n} < {(1 + \sqrt{2})}^{n - 1}$ $\frac{1}{2} {(\frac{2}{3})}^{n} < \frac{a_{n}}{3^{n}} < \frac{1}{(1 + \sqrt{2})} {(\frac{1 + \sqrt{2}}{3})}^{n}$ $\frac{1}{2} \underset{n = 1}{\sum^{\infty}} {(\frac{2}{3})}^{n} < \underset{n = 1}{\sum^{\infty}} \frac{a_{n}}{3^{n}} < \frac{1}{(1 + \sqrt{2})} \underset{n = 1}{\sum^{\infty}} {(\frac{1 + \sqrt{2}}{3})}^{n}$ $\frac{1}{2} \times \frac{2}{3 \times (1 - \frac{2}{3})} < \underset{n = 1}{\sum^{\infty}} \frac{a_{n}}{3^{n}} < \frac{1}{(1 + \sqrt{2})} \times \frac{1 + \sqrt{2}}{3 \times (1 - \frac{1 + \sqrt{2}}{3})}$ $1 < \underset{n = 1}{\sum^{\infty}} \frac{a_{n}}{3^{n}} < 1 + \frac{\sqrt{2}}{2} \approx 1.707$
	Commented by infinityaction last updated on 08/Oct/22

	$t h a n k s s i r$
	Commented by Tawa11 last updated on 09/Oct/22

	$Great sir .$
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