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Question:

Given $a_{n+1} = a_{n}+\sqrt{1+a^2_{n}}$ and $a_{0}=0$ and $a_{1} = 1,$ find $\displaystyle \lim_{n\rightarrow \infty}\left(\frac{a^2_{n}}{2^{n-1}}\right)$.

Attempt:

Assume $a_{n} = \cot \alpha_{n},$ then $\displaystyle \cot \alpha_{n+1} = \cot \alpha_{n}+\cos \alpha_{n} = \cot\frac{\alpha_{n}}{2}$.

Could someone help me?

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    Do you mean $\csc\alpha _n$ instead of $\cos\alpha_n$ in the second last step? The answer of your attempt is correct though. – 2017-01-31
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    There are a few posts on this site about $\lim a_n/2^n$. (I.e., without the square in the numerator.) See [this question](http://math.stackexchange.com/questions/1175041/convergence-of-sequence-given-by-x-1-1-and-x-n1-x-n-sqrtx-n21) and other posts [linked there](http://math.stackexchange.com/questions/linked/1175041). – 2017-01-31
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    @MartinSleziak Thank you for the extra information. The limit you mentioned is actually interesting. – 2017-01-31
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    are you sure you have copied the question properly.I had seen this question earlier in arihant trigonometry by amit m aggarwal .I donot have the book with me .You can perhaps consult the book it is a solved example there and answer your own question. – 2017-04-07

3 Answers 3

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Notice for all $a_{n\in\Bbb{N}}$ $$\frac{a_{n+1}}{a_n}=1+\sqrt{1+\frac1{a_n^2}}\ge2$$ $$\implies a_{n}=\prod^{n-1}_{k=1}\left(1+\sqrt{1+\frac1{a_n^2}}\right)\ge2^{n-1}$$

So the limit $$\lim_{n\to\infty}\frac{a^2_n}{2^{n-1}}\ge\lim_{n\to\infty}\frac{(2^{n-1})^2}{2^{n-1}}=\lim_{n\to\infty}2^{n-1}\to\infty$$

diverges.

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Intuitively, $a_n$ gets double every time. Therefore, $a_n^2$ gets multiplied by $4$. So you may expect the ratio to go to $\infty$, as $n \rightarrow \infty$, just because the nominator grows faster than the denominator.

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it's Easy to show $a_n$ be positive for all $n>0$ and $a_n$ is increasing sequence.

Then consider $\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}$ :

$$\frac{a_{n+1}}{a_n}=1+\sqrt{\frac{1}{a_n^2}+1}\ge2, \forall n\ge1$$

We have :

$$a_{n+1}\ge2a_n , \forall n\ge1$$

Therefore :

$$a_{n+1}\ge2^na_1=2^n$$

Next, I find the limit of $\frac{a_n^2}{2^{n-1}}$

$$\lim_{n\rightarrow\infty} \frac{a_n^2}{2^{n-1}}\ge\lim_{n\rightarrow\infty}\frac{(2^{n-1})^2}{2^{n-1}}=\infty$$