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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.

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(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.
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