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The sequence $(a_n)$ is defined by $a_1 = 1$ and $a_{n+1}=a_{n}+\sqrt{1+a_{n}^{2}}$. A sequence $(b_n)$ is defined by $b_n = \dfrac{a_n}{2^n}$. It is easy to see $(b_n)$ is monotone but is it convergent?

Suppose $(b_n)$ be bounded. Then $(b_n)$ is convergent by Monotone Convergence Theorem. Say, $b_n$ converges to a limit $b$, then $(b_{n+1}) \to b$.

We have $a_{n+1}=a_{n}+\sqrt{1+a_{n}^{2}}$. Dividing $2^{n+1}$ in both sides of the equation, $\Rightarrow \dfrac{a_{n+1}}{2^{n+1}}=\dfrac{a_{n}+\sqrt{1+a_{n}^{2}}}{2^{n+1}}$

$\Rightarrow b_{n+1} = \dfrac{b_n}{2} + \sqrt{\dfrac{1}{2^{2n+2}} + \dfrac{b_n^2}{4}}$

$\Rightarrow \lim_{n \to \infty} b_{n+1} = \lim_{n \to \infty} \dfrac{b_n}{2} + \lim_{n \to \infty} \sqrt{\dfrac{1}{2^{2n+2}} + \dfrac{b_n^2}{4}}$

$\Rightarrow b = \dfrac{b}{2} + \sqrt{0 + \dfrac{b^2}{4}}$

$\Rightarrow b = b$.

No decisive conclusions could be made.

However numerical evidences suggest that $ b_n \approx 0.63$ whenever $100\le n \le 200$. Does it diverge to infinity but sure, it diverges pretty slow? I tried to prove otherwise that $b_n$ is bounded above by $1$ but it didn't work but neither can find an counterexample to it. How far do I need to go in the sequence to surpass $1$?

ash
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2 Answers2

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Using the binomial theorem gives $$a_{n+1}=2a_n+\frac1{2a_n}+O(a_n^{-3})=2a_n+O(1/a_n).$$ Therefore $$b_{n+1}=b_n+O\left(\frac1{2^{2n}b_n}\right).$$ As $(b_n)$ is increasing, $\sum_n\frac1{2^{2n} b_n}$ is convergent. Therefore $(b_n)$ is bounded.

Angina Seng
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$$ b_{n+1} = \dfrac{b_n}{2} + \sqrt{\dfrac{1}{2^{2n+2}} + \dfrac{b_n^2}{4}} $$ where $b_1 = 1/2$. If you try to bound the relation, you get $$ \dfrac{b_n}{2} + \sqrt{\dfrac{1}{2^{2n+2}} + \dfrac{b_n^2}{4}} \le b_n + x_n $$$$ \dfrac{1}{2^{2n+2}} \le b_nx_n + x_n^2 $$ so $x_n = 2^{-n}$ is enough. This proves that $$b_{n+1}\le b_n + 2^{-n}\le b_1 + 1$$ so the sequence is bounded, and the monotone sequence converges.

Exodd
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