Evaluate:$$\displaystyle \lim_{x \rightarrow 0}\frac{\cos(1-\frac{\sin x}{x})+\cos(2-\frac{\sin(2x)}{x})+\cdots+\cos(k-\frac{\sin(kx)}{x})-k}{\cos(1-\cos(x))+\cos(2-2\cos(2x))+\cdots+\cos(k-k\cos(kx))-k}$$

I can find this limit using L' Hospital Rule, I do not know how to do it without that.This question is proposed to Romanian Math Magazine by Jalil Hajimir

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    $\begingroup$ I think there is a typo here $\cos(k-\color{red}k\cos(kx))$ $\endgroup$ – user Dec 4 '19 at 14:30
  • $\begingroup$ I fixed it. Thanks $\endgroup$ – Keon Dec 4 '19 at 15:00
  • $\begingroup$ Can you give a more precise reference (including a link, if there is one)? $\endgroup$ – Barry Cipra Dec 4 '19 at 15:17

Let's observe that $$\frac{1-\cos(j-j\cos jx)} {j^2(1-\cos jx) ^2}\to\frac{1}{2}$$ and hence $$\frac{1-\cos(j-\cos jx)} {x^4}=\frac{1-\cos(j-\cos jx)} {j^2(1-\cos jx) ^2}\cdot j^2\cdot\frac{(1-\cos jx)^2} {(jx)^4}\cdot j^4\to \frac{j^6}{8}$$ Thus if we change sign of both numerator and denominator of the expression under limit in question and divide both numerator and denominator by $x^4$ then the denominator of resulting expression tends to $\sum_{j=1}^{k}j^6/8$.

To deal with numerator let's observe that a typical term of numerator is $$\dfrac{1-\cos\left(j-\dfrac{\sin jx} {x} \right)} {x^4}=\dfrac{1-\cos\left(j-\dfrac{\sin jx} {x} \right)} {\left(j-\dfrac{\sin jx} {x} \right) ^2}\cdot\frac{(jx-\sin jx) ^2} {(jx) ^6}\cdot j^6\to\frac{j^6}{72}$$ and hence the numerator tends to $\sum_{j=1}^{k}j^6/72$. The desired limit is thus $1/9$. Note that we need here the result that $$\frac{x-\sin x} {x^3}\to\frac{1}{6}$$ which is easily handled via Taylor series.


Let proceed by Taylor’s series to obtain






$$\frac{\sum_{i=1}^k \frac{i^6x^4}{72}+o(x^4)}{\sum _{i=1}^k \frac{i^6x^4}{8}+o(x^4)}= \frac{\sum_{i=1}^k \frac{i^6}{72}+o(1)}{\sum _{i=1}^k \frac{i^6}{8}+o(1)}\to \frac19 \frac{\sum_{i=1}^k i^6}{\sum _{i=1}^k i^6}=\frac19$$


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