# If $a_1=1,a_{n+1}^2-2a_na_{n+1}-a_n=0$ then sum of the series $\sum {a_n\over 3^n}$ [duplicate]

Let $$\{a_n\}$$ be a sequence of positive real numbers such that $$a_1=1,a_{n+1}^2-2a_na_{n+1}-a_n=0\,\,\forall n\ge 1$$. Then show that $$1<\sum_{n=1}^\infty {a_n\over 3^n}<2$$. I got this problem from a question paper. Though it didn't asked for exact sum of the series , if you also manage to find that then please post it.

(Fill in the gaps as needed. If you're stuck, show your work and explain what you've tried.)

Hints:

• After calculating the first few terms, the ratio of the terms is very similar. If we estimate $$a_n \approx a_1 r^{n-1}$$, this suggests we should solve for $$\frac{ 1/3} { 1 - (r_1/3) } = 1$$ and $$\frac{1/3} { 1 - (r_2/3)} = 2$$ to give us an idea of how to bound the sequence. This gives us $$r_1 = 2, r_2 = 5/2$$, so we want to show that $$2 a_n < a_{n+1} < \frac{5}{2} a_n$$ (with some flexibility if this doesn't immediately work out).
• Show that $$a_{n+1} = \frac{ 2a_n + \sqrt{ 4a_n^2 + 4a_n } } { 2}$$. In particular, reject the negative root.
• Show that $$a_n \geq 1$$.
• Show that $$2 a_n < a_{n+1} < \frac{5}{2} a_n$$.
• Hence, show that $$1=\frac{ 1/3 } { 1 - 2/3} < \sum \frac{a_n}{3^n} < \frac{ 1 / 3 } { 1 - 5/6}=2$$

Note:

• The LHS is true by calculating the first 5 terms.
• Of course, we can't prove the RHS just by calculating enough terms.
• In fact, the bounding inequality $$2a_n < a_{n+1} < 2a_n + \frac{1}{2}$$, so the ratio $$a_{n+1} / a_n$$ is very close to 2, esp at (slightly) larger values of $$n$$.
• Not surprisingly, the value of the summation is much much closer to the 1. Using the first ~10 terms to get a better approximation, you can in fact show that the summation is between 1.2 and 1.25.