# Then sum of series $\sum_{n=1}^{\infty} \frac{a_n}{3^n}$ lies in the interval

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?

• What have you tried? – Robert Z Feb 25 '19 at 15:07
• @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 – Eklavya Feb 25 '19 at 15:09
• So can you use that to approximate $a_n$? You aren't looking for an exact sum, after all. – rogerl Feb 25 '19 at 15:12

Yes, $$a_{n+1}=a_n +\sqrt{a_n^2+a_n}$$ and therefore for $$n\geq 1$$, $$2a_{n}=a_n +\sqrt{a_n^2} which implies, together with $$a_1=1$$, that for $$n>1$$, $$2^{n-1}< a_n\leq (1+\sqrt{2})^{n-1}.$$ Can you take it from here?
• Doesn't this assume that ${a_n}^2 \geq a_n$? What if $a_n<1$? – Akash Gaur Mar 21 '19 at 8:55
• @Rhaldryn Note that $a_1=1$ and $a_{n+1}=a_n +\sqrt{({a_n}^2+a_n)}>a_n$. So $a_n^2\geq a_n$. – Robert Z Mar 21 '19 at 9:55