Find the limit $\lim_{n\to \infty} k_n/2^n$ for $k_1=0$ and $k_{n+1}=k_n+\sqrt{1+k_n^2}$

Question

$k_n$ 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.

Answer 1

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.

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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: $\displaystyle \sum_{i=1}^n 2^{-i}=1-2^{-n}$

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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 $|\sec{x}|=\sec{x}$.

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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.

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So finally, $\displaystyle \lim_{n \to \infty} \frac{k_n}{2^n}=\frac{2}{\pi}$ .

Answer 2

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.