Given $a_{n+1} = a_{n}+\sqrt{1+a^2_{n}}$ and $a_{0}=0$ and $a_{1} = 1$, find $\lim_{n\to\infty}a^2_{n}/2^{n-1}$

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 $\frac{a_n^2}{2^{n-1}}$

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