# computing a trignometric limit

I am trying to show that $$\lim _{n\rightarrow \infty }\dfrac {1+\cos \dfrac {x} {n}+\cos \dfrac {2x} {n}+\ldots +\cos\dfrac {\left( n-1\right) x} {n}} {n } = \dfrac{\sin x}{x}$$

I have attempted a number of approaches such as trigonometric tricks and the most recent with substitution of $\cos \dfrac {x} {n}=\dfrac {e^{i\frac {x} {n}}+e^{-\frac {ix} {n}}} {2}$ Starting from the left hand side i ended up with an expression such as $$\lim _{n\rightarrow \infty }\frac {\frac {1-e^{ix}} {1-e^{\frac {ix} {n}}}+\frac {1-e^{-ix}} {1-e^{\frac {-ix} {n}}}} {2n}$$ I am unsure how i could possibly rearrange this so upon taking the limit i can convert to RHS.

Any help with this method or even an alternative proof strategy would be much appreciated.

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For a "pre-integral calculus" method:

You can use the formula

$$\sum_{k=0}^{n-1} \cos k d = \frac{ \sin (nd/2)}{\sin (d/2)} \cos ((n-1)d/2)$$

Set $\displaystyle d = \frac{x}{n}$ and use the limit $\displaystyle \lim_{t\to 0} \frac{\sin t}{t} = 1$ and the formula $\displaystyle 2 \sin (x/2) \cos (x/2) = \sin x$.

For a proof of the formula, see here: How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?

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My alternate answer follows closely to Hans Lundmark's "answer" to your cited post. – robjohn Apr 2 '12 at 20:26
@robjohn: Indeed! +1. – Aryabhata Apr 2 '12 at 20:28

You can compute the sum or recognize a Riemann sum: for $x\neq 0$ $$\lim_{n\to+\infty}\frac 1n\sum_{k=0}^{n-1}\cos\frac{kx}n=\int_0^1\cos(tx)dt=\left[\frac{\sin(tx)}x\right]_{t=0}^{t=1}=\frac{\sin x}x.$$

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firstly thank you for your answer i suspected such an answer from one of your other posts, but i have n't studied riemann sums yet so was not sure. I Have just one question how did you calculate the upper and lower limits of the integral sign ? is that because we have k = 0 and then k nearly equal to infinity/infinity in the previous step. – Hardy Apr 2 '12 at 19:42
+1: I was about to mention this in my answer too! – Aryabhata Apr 2 '12 at 19:44
@Hardy: The terms of the sum are taken at intervals of $x/n$. This corresponds either to a Riemann sum for the function $\cos tx$ over the interval $0\le t\le 1$ and intervals of width $1/n$, or to a Riemann sum for $\cos t$ over the interval $0\le t\le x$, with intervals of width $x/n$. Davide took the first approach; the second yields $$\frac1x\lim_{n\to\infty}\frac{x}n\sum_{k=0}^{n-1}\cos\left(k\frac{x}n\right)=\‌​frac1x\int_0^x\cos t dt=\frac1x[\sin t]_0^x=\frac{\sin x}x\;.$$ – Brian M. Scott Apr 2 '12 at 19:52
Beat me to it (+1) – robjohn Apr 2 '12 at 19:53
@BrianM.Scott Thank you very much for that. I think i understand it much better now. – Hardy Apr 2 '12 at 19:56

Your sum is a Riemann sum approximating the integral $$\int_0^1\cos(tx)\,\mathrm{d}t=\frac1x\int_0^x\cos(t)\,\mathrm{d}t=\frac{\sin(x)}{x}$$ Alternatively, your sum is the real part of the geometric sum $$\newcommand{\cis}{\operatorname{cis}} \frac1n\left(1+\cis\left(1\frac{x}{n}\right)+\cis\left(2\frac{x}{n}\right)+\dots+\cis\left((n-1)\frac{x}{n}\right)\right) =\frac{\cis(x)-1}{n\left(\cis\left(\frac{x}{n}\right)-1\right)}\tag{1}$$ where $\cis(x)=e^{ix}=\cos(x)+i\sin(x)$.

Then, note that \begin{align} \lim_{n\to\infty}n\left(\cis\left(\frac{x}{n}\right)-1\right) &=\lim_{n\to\infty}n\left(\cis\left(\frac{x}{2n}\right)-\cis\left(-\frac{x}{2n}\right)\right)\cis\left(\frac{x}{2n}\right)\\ &=\lim_{n\to\infty}2in\sin\left(\frac{x}{2n}\right)\cis\left(\frac{x}{2n}\right)\\ &=ix\tag{2} \end{align} Combining $(1)$ and $(2)$ yields $$\lim_{n\to\infty}\frac1n\left(1+\cis\left(1\frac{x}{n}\right)+\cis\left(2\frac{x}{n}\right)+\dots+\cis\left((n-1)\frac{x}{n}\right)\right)=\frac{\cis(x)-1}{ix}\tag{3}$$ Taking the real parts of $(3)$ yields $$\lim_{n\to\infty}\frac1n\left(1+\cos\left(1\frac{x}{n}\right)+\cos\left(2\frac{x}{n}\right)+\dots+\cos\left((n-1)\frac{x}{n}\right)\right)=\frac{\sin(x)}{x}\tag{4}$$

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what is cis or was it meant to be cos ? – Hardy Apr 2 '12 at 20:47
@Hardy: I'm sorry; I forget that the function is not as widespread as I assume. $\cis(x)=e^{ix}=\cos(x)+i\sin(x)$. I have amended my post. – robjohn Apr 2 '12 at 20:50
Thanks, also your answer is bit advanced, but nicely done. Thanks for sharing – Hardy Apr 2 '12 at 21:03
@Hardy: actually, I am really only proving the identity that Aryabhata cites, but I do use complex functions. However, that is why I kept my first answer using Riemann Sums. Even though it was already given by Davide Giraudo, it is a lot simpler. – robjohn Apr 2 '12 at 21:16

Here is slightly long winded solution using trigonometric identities only. $$\lim _{n\rightarrow \infty }\dfrac {1+\cos \dfrac {x} {n}+\cos \dfrac {2x} {n}+\ldots +\cos\dfrac {\left( n-1\right) x} {n}} {n }$$

using the trignometric identity $$\sum _{n=1}^{N}\cos n\theta =\dfrac {-1} {2}+\dfrac {\sin \left( N+\dfrac {1} {2}\right) \theta } {2\sin \dfrac {\theta } {2}}$$ our equation can be rewritten as $$\lim _{n\rightarrow \infty }\dfrac {1-\dfrac {1} {2}+\dfrac {\sin \left( n-1+\dfrac {1} {2}\right) \dfrac {x} {n}} {2\sin x / 2n}} {n}$$

$$\lim _{n\rightarrow \infty }\dfrac {1} {2n}\left( \dfrac {\sin\dfrac {x} {2n}+\sin \left( x-\dfrac {x} {2n}\right) } {\sin \dfrac {x} {2n}}\right)$$

$$\lim _{n\rightarrow \infty }\dfrac {1} {2n}\left( \dfrac {2\sin\dfrac {x} {2}\cos \left(\dfrac {\dfrac {x} {n}-x} {2}\right)} {\sin \dfrac {x} {2n}}\right)$$

The following restructure of the denominator was pointed out to me by Aryabhata. :-) $$\lim _{n\rightarrow \infty }\dfrac {\sin \dfrac {x} {2}\cos \dfrac {x} {2}\left( \dfrac {1} {n}-1\right) } {\dfrac {x\sin \dfrac {x} {2n}} {2\dfrac {x} {2n}}}$$

Taking the limit we get $$\dfrac {\sin \dfrac {x} {2}\cos -\dfrac {x} {2}} {\dfrac {x} {2}}$$ which is the same as $$\dfrac {\sin \dfrac {x} {2}\cos \dfrac {x} {2}} {\dfrac {x} {2}}$$ which gives RHS using the trigonometric identity once again. $$\dfrac{\sin x}{x}$$

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