$\{a_n\}$ be a sequence such that $ a_{n+1}^2-2a_na_{n+1}-a_n=0$, then $\sum_1^{\infty}\frac{a_n}{3^n}$ lies in... Let $\{a_n\}$ be a sequence of positive real numbers such that 
$a_1 =1,\ \  a_{n+1}^2-2a_na_{n+1}-a_n=0, \ \ \forall n\geq 1$. 
Then the sum of the series $\sum_1^{\infty}\frac{a_n}{3^n}$ lies in...
(A) $(1,2]$, (B) $(2,3]$, (C) $(3,4]$, (D)$(4,5]$. 
Solution attempt: 
Firstly, we figure out what $\frac{a_{n+1}}{a_n}$ is going to look like. We get, from the recursive formula, $\frac{a_{n+1}}{a_n}=1+\sqrt{1+\frac{1}{a_n^2}}$ (remembering the fact that $a_n>0$, the other root is rejected). 
We know that, if $\lim_{n \to \infty}\frac{a_{n+1}}{a_n}>1$, then $\lim a_n \to \infty$. 
Further, $(a_{n+1}-a_n)= \sqrt{a_n(a_n+1)}>0$. (Again, the other root is rejected due to the same reason). 
Hence, $(a_n)$ increases monotonically. 
Therefore, the largest value of $\frac{a_{n+1}}{a_n}$ is approximately $1+\sqrt{1+\frac{1}{1}} \approx 2.15$
Now, the sum can be approximated as $\displaystyle\frac{\frac{1}{3}}{1-\frac{2.15}{3}} \approx 1.3$ (In actuality, $\mathbb{sum}< 1.3$).
So, option $(A)$ is the correct answer. 
Is the procedure correct? 
I have been noticing a handful of this type of questions (based on approximations) lately, and the goal is to find out where the sum / the limit of the sequence might lie. 
Is there any "definitive" approach that exploits the recursive formula and gives us the value, or does the approach varies from problem to problem?

$a_{n+1}=a_n+\sqrt{a_n^2+a_n}>2a_n\forall n\ge 1$ $a_1=2^0,a_2>2,a_3>2^2,...,a_n\ge 2^{n-1}$ $1\le a_n\Rightarrow a_n+a_n^2\le 2a_n^2\Rightarrow a_{n+1}\le (\sqrt 2+1)a_n$ $a_1=(\sqrt 2+1)^0,a_2=(\sqrt 2+1),a_3<(\sqrt 2+1)^2,...,a_n\le (\sqrt 2+1)^{n-1}$ $\therefore 2^{n-1}\le a_n \le (\sqrt 2+1)^{n-1}$ $\Rightarrow \frac 13(\frac 23)^{n-1}\le {a_n\over 3^n}\le \frac 13({\sqrt 2+1 \over 3})^{n-1}$ $\therefore \frac 13\sum (\frac 23)^{n-1} < \sum {a_n\over 3^n}\le \frac 13 \sum ({\sqrt 2+1 \over 3})^{n-1}$
$\text{Hence}\,\,1<\sum {a_n\over 3^n}\le {1\over 2-\sqrt 2}\approx 1.707$
