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Let $(a_n)_n$ be sequence of positive real numbers such that $$a_1=1, \, a_{n+1}^2 -2a_na_{n+1}-a_n=0$$ for all $n \geq1$ Then sum of series $\sum_{n=1}^{\infty} \frac{a_n}{3^n}$ lies in the interval

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

My work. I found that $a_{n+1}=a_n +\sqrt{({a_n}^2+a_n)}$ then i tried to put this value in summation , but got stuck because of root .any suggestion?

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  • $\begingroup$ What have you tried? $\endgroup$ – Robert Z Feb 25 '19 at 15:07
  • $\begingroup$ @Robert $a_{n+1}=a_n +\sqrt{({a_n}^2+a_n)}$ then i tried to put this value in summation , but got stuck because of root .any suggestion $\endgroup$ – Eklavya Feb 25 '19 at 15:09
  • $\begingroup$ So can you use that to approximate $a_n$? You aren't looking for an exact sum, after all. $\endgroup$ – rogerl Feb 25 '19 at 15:12
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Yes, $a_{n+1}=a_n +\sqrt{a_n^2+a_n}$ and therefore for $n\geq 1$, $$2a_{n}=a_n +\sqrt{a_n^2}<a_{n+1}\leq a_n +\sqrt{a_n^2+a_n^2}=(1+\sqrt{2})a_{n}$$ which implies, together with $a_1=1$, that for $n>1$, $$2^{n-1}< a_n\leq (1+\sqrt{2})^{n-1}.$$ Can you take it from here?

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  • $\begingroup$ ya .i got it .answer is option A. Thanks $\endgroup$ – Eklavya Feb 25 '19 at 15:25
  • $\begingroup$ Yes, you are correct. $\endgroup$ – Robert Z Feb 25 '19 at 15:26
  • $\begingroup$ Doesn't this assume that ${a_n}^2 \geq a_n$? What if $a_n<1$? $\endgroup$ – Akash Gaur Mar 21 '19 at 8:55
  • $\begingroup$ Also can we find the exact sum? $\endgroup$ – Akash Gaur Mar 21 '19 at 9:17
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    $\begingroup$ @Rhaldryn Note that $a_1=1$ and $a_{n+1}=a_n +\sqrt{({a_n}^2+a_n)}>a_n$. So $a_n^2\geq a_n$. $\endgroup$ – Robert Z Mar 21 '19 at 9:55

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