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What is the limit of $$\lim_{x\to0} \frac{\cos x + \cos 2x + \dots+ \cos nx - n}{\sin x^2}$$

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closed as off-topic by Bruno Joyal, Stefan4024, user1337, Old John, Thomas Dec 17 '13 at 19:20

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  • $\begingroup$ Perhaps this link might come in handy here for computing the sum(s). $\endgroup$ – Lucian Dec 17 '13 at 18:30
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    $\begingroup$ @Lain : I suspect the reason for the votes to close is that you haven't included any thoughts beyond the question, and it's phrased in a way suitable for homework problems. $\endgroup$ – Michael Hardy Dec 17 '13 at 18:56
  • $\begingroup$ Oh, sorry, I'm a newbie here, I don't know all the rules. Thank you! I've solved it, thank you to you all. $\endgroup$ – Lain Dec 17 '13 at 19:04
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$$ \begin{align} & \lim_{x\to0}\frac{\cos x+\cos 2x+\cdots+\cos nx-n}{\sin x^2} \\[10pt] & =\lim_{x\to0} \frac{\cos x -1+\cos 2x -1+\cdots+\cos nx-1}{x^2} \cdot\frac{x^2}{\sin x^2} \\[10pt] & = -\left(\frac{1}{2} + \frac{4}{2}+\frac{9}{2}+\cdots+\frac{n^2}{2}\right)\cdot1=-\frac{n(n+1)(2n+1)}{12}. \end{align} $$ We applied $$\lim_{x\to0}\frac{1-\cos nx}{x^2}=\frac{n^2}{2} $$

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We have by taylor series $$\cos(kx)\sim_0 1-\frac{k^2}{2}x^2$$ and $$\sin x^2\sim_0 x^2$$ hence $$\lim_{x\to0} \frac{(\cos x + \cos 2x + ...+ \cos nx - n)}{\sin x^2}= -\frac{1}{2}\sum_{k=1}^n k^2=-\frac{n(n+1)(2n+1)}{12}$$

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  • $\begingroup$ $\ddot\smile! +1$ $\endgroup$ – Namaste Dec 18 '13 at 14:33
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We have $$\sum_{k=1}^n \cos(kx) = \dfrac{\sin(nx/2) \cos((n+1)x/2)}{\sin(x/2)}$$ Hence, we have $$\dfrac{\dfrac{\sin(nx/2) \cos((n+1)x/2)}{\sin(x/2)} - n}{\sin(x^2)} = \dfrac{\sin(nx/2) \cos((n+1)x/2) - n\sin(x/2)}{\sin(x^2)\sin(x/2)}$$ Expanding around $0$ gives us $$\dfrac{\left(nx/2 - \dfrac{(nx/2)^3}{3!} + \mathcal{O}(x^5)\right) \left(1-\dfrac{(n+1)^2x^2/4}{2!} + \mathcal{O}(x^4)\right)-n\left(x/2 - \dfrac{(x/2)^3}{3!} + \mathcal{O}(x^5) \right)}{x^3/2 + \mathcal{O}(x^4)}$$ Now cancel off the terms and conclude the limit.

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  • $\begingroup$ What is $\mathcal O$? $\endgroup$ – JMCF125 Dec 17 '13 at 18:59
  • $\begingroup$ @JMCF125 You may want to look here $\endgroup$ – user17762 Dec 17 '13 at 19:07
  • $\begingroup$ Ah, I'm used to seeing it as $O$ or $\Theta$. The page doesn't refer that notation. $\endgroup$ – JMCF125 Dec 17 '13 at 19:09
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As $x \to 0$ both the numerator and denominator tend to zero, so you can apply L'Hôpital's rule to get $$\lim_{x \to 0} \dfrac{\sum_{k=1}^n \cos kx - n}{\sin x^2} = \lim_{x \to 0} \dfrac{\sum_{k=1}^n -k\sin kx}{2x\cos x^2} = \lim_{x \to 0} \dfrac{-1}{2\cos x^2} \sum_{k=1}^n \dfrac{k\sin kx}{x}$$

Can you see what to do now?

Hint (hover mouse over to see):

Write $$\dfrac{k \sin kx}{x} = k^2 \dfrac{\sin kx}{kx}$$ and use the fact that $$\lim_{\theta \to 0} \dfrac{\sin \theta}{\theta} = 1$$

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Try using the L'Hopital Rule. Since the numerator and denominator both tend to 0 as x tends to 0, the limit of the quotient will be the same as the limit of the quotient of the differentiated functions.

$$\lim_{x\to0} \frac{cosx+cos2x+...+cosnx -n}{sinx^2}=\lim_{x\to0} \frac{-sinx-2sin2x-...-nsinnx}{cosx^2*2x}=\lim_{x\to0} \frac{-cosx-4cos2x-...-n^2cosnx}{2cosx^2-2xsinx^2}$$ (Applying L'Hopital Rule twice)

This limit is minus half of the sum of squares from 1 to n.

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  • $\begingroup$ Typed too slowly as I'm a beginner in these type of texts. I didn't realize someone posted the same answer as mine. :( $\endgroup$ – ireallydonknow Dec 17 '13 at 18:43
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    $\begingroup$ You can right click on others' Latex to see how they wrote it. For common functions you can write \sin and \cos instead of sin and cos, and use \cdot for explicit multiplication if you like. $\endgroup$ – DanielV Dec 17 '13 at 19:08

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