Find the limit $\lim_{n\to \infty} k_n/2^n$ for $k_1=0$ and $k_{n+1}=k_n+\sqrt{1+k_n^2}$ $k_n$ is defined with $k_1=0$ and $k_{n+1}=k_n+\sqrt{1+k_n^2}$.
This is homework, please do not provide a complete solution
edit : One of the many things I tried is to make it the root of an equation like this . I first found that and then tried $k_{n+1}=\sqrt{1+2 k_n k_{n+1}}$ then $(\lim_{n\to \infty} k_{n+1}/2^{n+1})^2=\lim_{n\to \infty}  (1+2 k_n k_{n+1})/2^{2n+2}=\lim_{n\to \infty}  k_n k_{n+1}/2^{2n+1}$ but then I was stuck and found nothing to continue. All my other trials ended up stuck.
 A: You need to think just a little bit outside the box for this one and not bruteforce your way through to a solution. The first thing that jumped at me was the $\sqrt{1+k^2}$, which made me think that either the hyperbolic sine and cosine are going to be useful here, or it's the tangent and secant. The latter was the right approach.

First, two elementary lemmas we'll use:

Lemma 1: $\displaystyle \tan{x}+\sec{x}=\tan{\left( \frac{x}{2}+\frac{\pi}{4}\right)}$
Lemma 2: $\displaystyle \sum_{i=1}^n 2^{-i}=1-2^{-n}$


Now, we have: 


*

*$k_0=\tan{0}=0$

*$k_1=\tan{0}+\sqrt{1+\tan{0}^2}=\tan{0}+|\sec{0}|=\tan{0}+\sec{0}= \tan{(0/2+\pi/4)}$

*$k_2=\tan{\pi/4}+\sec{\pi/4}=\tan{\left(\frac{\pi}{2}\left(\frac{1}{2}+\frac{1}{4} \right)\right)}$

*...

*$k_n=\tan{\left(\frac{\pi}{2}\left(\frac{1}{2}+\frac{1}{4} \cdots \frac{1}{2^n} \right)\right)}=\tan{\left(\frac{\pi}{2}\left(1-2^{-n} \right)\right)}$


The general form above is only true because for $0 \leq x < \pi/2$, we have $|\sec{x}|=\sec{x}$.

So now, we calculate our limit:
\begin{eqnarray*}
\displaystyle \lim_{n\to \infty} \frac{k_n}{2^n} &=& \lim_{n\to \infty} 2^{-n} \tan{\left(\frac{\pi}{2}\left(1-2^{-n} \right)\right)} \\ &=& \lim_{x\to 0} x \tan{\left(\frac{\pi}{2}\left(1-x \right)\right)}\\
&=& \left(  \lim_{x\to 0} \sin{\frac{\pi}{2}(1-x)} \right) \cdot \left( \lim_{x\to 0} \frac{x}{\cos{\frac{\pi}{2}(1-x)}} \right)\\
&=& 1 \cdot  \lim_{x\to 0} \frac{\frac{\mathrm{d}}{\mathrm{d} x}x}{\frac{\mathrm{d}}{\mathrm{d} x}\cos{\frac{\pi}{2}(1-x)}} \\
&=& \lim_{x\to 0} \frac{1}{\frac{\pi}{2} \sin{\frac{\pi}{2}(1-x)}}\\
&=& \frac{1}{\frac{\pi}{2} \sin{\frac{\pi}{2}(1-0)}}\\&=& \boxed{\frac{2}{\pi}}
\end{eqnarray*}
And of course, the fourth equality is justified by l'Hôpital's rule.

So finally, $\displaystyle \lim_{n \to \infty} \frac{k_n}{2^n}=\frac{2}{\pi}$ .
A: suppose $k_n$ is $cotø$ then $$k_{n+1}=cotø+cosecø$=$(1+cosø)/sinø=2cos^(ø/2)/2sin(ø/2)cos(ø/2)=cot(ø/2)$$
now as $k_0=0$ take $k_0=cotπ/2$ therefore $k_n=cot(π/2^{n+1})$ now i guess you can solve the limit.the denominator will have $0*∞$ form where $2^n$ cancels.
