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$k_n$ 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.

Martin Sleziak
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user178056
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  • Can you show us what you've done so far? – Robin Goodfellow Sep 23 '14 at 00:18
  • Why the thumbs down? Robin: I tried a lot of stuff but none gave me any progress. I tried many ways to calculate $k_{n+1}/2^{n+1}$ from $k_n/2^n$ but none were useful for me. I don't understand the thumbs downs. I did something wrong? – user178056 Sep 23 '14 at 00:23
  • The downvotes are because you didn't demonstrate any effort in asking your question. For instance, you haven't told us what you've already tried. – Robin Goodfellow Sep 23 '14 at 00:27
  • Robin: I'm sorry, I didn't know it was needed because nothing interesting came out of it. I really have no idea how to approach this problem, and all the things I tried doing ended up completely useless. – user178056 Sep 23 '14 at 00:28
  • You could, at least, write down what you have tried. – Robin Goodfellow Sep 23 '14 at 00:31
  • Robin: okay, here – user178056 Sep 23 '14 at 00:40
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    Some may be downvoting because of the sentence "This is homework, please be complete in your explanation", which many would interpret as "Do my homework for me". I think you would get a much better response by changing it to "This is homework, please do not provide a complete solution". – Antonio Vargas Sep 23 '14 at 00:46
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    Antonio: Oh, really really sorry. I realise it sounds cocky now. I will change it. – user178056 Sep 23 '14 at 00:50
  • Now the question is surely deserving of an upvote :) – Antonio Vargas Sep 23 '14 at 00:51
  • Much better now – Aaron Maroja Sep 23 '14 at 00:56
  • Here are two comments to get you started: First, the limit, which you can compute numerically, is $1/\pi$. Second, the limit depends on the initial value $k_1 = 0$. – Yuval Filmus Sep 23 '14 at 00:58
  • Should I just accept the answer now or wait for more answers? I saw someone saying to always wait in another question and I don't want to mess up again by accepting too fast. – user178056 Sep 23 '14 at 01:35
  • See also http://math.stackexchange.com/questions/1333944/how-to-evaluate-lim-n-to-infty-fraca-n2n-1-if-a-0-0-and-a-n and http://math.stackexchange.com/questions/1175041/convergence-of-sequence-given-by-x-1-1-and-x-n1-x-n-sqrtx-n21 – Martin Sleziak Jan 31 '17 at 12:39

2 Answers2

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

Fujoyaki
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  • This answer made me happy. –  Sep 23 '14 at 01:10
  • Thank you sir. I still have to prove the lemma and go on this step by step, but the result 2/pi agrees with my computations, so I don't doubt the answer. – user178056 Sep 23 '14 at 01:19
  • @ChantryCargill Thank you :)

    @ user178056 If anything is unclear, you can ask me in the comments.

     – Fujoyaki Sep 23 '14 at 01:26
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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.

avz2611
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