Find limit n tends to infinity

Question

$a_{n+1} = a_{n} +\sqrt{1+a^2_{n}} , a_{1} = 0, \lim_{x \to \infty} \frac{a_{n}}{ 2^{n-1}}$ I tried it by using recurrence relation but couldn't proceed further. Also I tried to summation to cancel out .

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MathJax is just a TeX based markup language for typesetting Mathematics. See the links that I provided in my previous comment. As an example, $a_{n+1}$ may be typeset using $a_{n+1}$ . Consider reviewing the edit that I proposed. — Brian61354270, Mar 24 '19 at 17:36
I am very sorry @st.math. I am using it for first time . I didn't even know when I rejected it . I may have done it by mistake. Can I guide me please — Kartikeya Kansal, Mar 24 '19 at 17:52
Possible duplicate of Limit $\lim_{n \to \infty} \frac {a_n}{2^{n-1}}=\frac 4{\pi}$ for $a_{n+1}=a_n+\sqrt{1+a_n^2}$ and $a_0=0$ — Sil, Mar 24 '19 at 18:17
Here is a website to help you write latex: http://detexify.kirelabs.org/classify.html — user614287, Mar 24 '19 at 18:24

score 1 · Accepted Answer · answered Mar 24 '19 at 18:00

The key is recognizing the recurrence formula $$ a_{n+1} = a_n + \sqrt{1+a_n^2} $$ as a disguise for the half-angle formula for cotangent: $$ \cot\frac{x}{2} = \cot x + \csc x = \cot x + \sqrt{1 + \cot^2 x}. $$

Here, $a_1 = 0 = \cot \dfrac{\pi}{2}$, so $a_2 = 1 = \cot \dfrac{\pi}{4}$, and in general $a_n = \cot \dfrac{\pi}{2^n}$.

Using the fact that $\lim_{x\to 0} x\cot x = 1$, $$ \lim_{n\to\infty} \frac{a_n}{2^{n-1}} = \lim_{n\to\infty}\frac{1}{2^{n-1}}\frac{2^n}{\pi}\bigg(\frac{\pi}{2^n}\cot \frac{\pi}{2^n}\bigg) = \frac{2}{\pi}. $$

Thanks I tried it by substituting it with tanx but then it became tan(π/4 - x) then I left it there only. — Kartikeya Kansal, Mar 24 '19 at 18:05

Find limit n tends to infinity

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