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What is the value of $$\displaystyle{\lim_{n\to \infty}\frac{1}{n}\sum_{r=0}^{n-1}\cos\frac{r\pi}{2n}}$$

The answer is given as $1$ but I am almost 100% sure it will not be $1$. What will be the correct answer and how to find it.

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    $\begingroup$ "I am almost 100% sure it will not be 1": what makes you say that ? By the way, this is a definite integral in disguise. $\endgroup$ – Yves Daoust Jun 25 '14 at 9:24
  • $\begingroup$ is it $x \to \infty$ or $n \to \infty$?? $\endgroup$ – tattwamasi amrutam Jun 25 '14 at 9:24
  • $\begingroup$ @YvesDaoust : As n terms have been added and each term is less than or equal to 1 so sum is less than n $\endgroup$ – User Not Found Jun 25 '14 at 9:25
  • $\begingroup$ @Arghya: quite right. Do you see the integral ? $\endgroup$ – Yves Daoust Jun 25 '14 at 9:27
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    $\begingroup$ @ArghyaChakraborty Have you learned integration yet? If so, try to see a Riemann sum in that sum. And if you do, you will find the limit is $\frac{2}{\pi}$. $\endgroup$ – Daniel Fischer Jun 25 '14 at 9:29
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Hint : This is equivalent to the integral $\int_{0}^1 \cos {\dfrac{\pi x}{2}}dx $

In general, $$\int_{0}^1f(x)dx=\lim_{n\rightarrow \infty}\sum_{r=0}^{n-1}\dfrac{1}{n}f(\dfrac{r}{n})$$

Also,

$$\int_{a}^bf(x)dx=\lim_{n\rightarrow \infty}\sum_{r=0}^{n-1}\dfrac{b-a}{n}f(a+r\dfrac{b-a}{n})$$

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  • $\begingroup$ @DanielFischer .. Thanks .. I have edited accordingly. Actually I was in a hurry to answer the question :D $\endgroup$ – Debashish Jun 25 '14 at 9:35
  • $\begingroup$ @Arghya Chakraborty .. Refer any Higher secondary book on Integration for more details. Or if you have some doubts you may ask. $\endgroup$ – Debashish Jun 25 '14 at 9:43
  • $\begingroup$ Thanks, That is exactly what I needed. $\endgroup$ – User Not Found Jun 25 '14 at 9:43
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Without integrals, using complex numbers.

Let $2\theta=\frac{\pi}{2n}$ and sum the geometric progression:

$$\sum_{r=0}^{n-1}e^{2ir\theta}=\frac{e^{2in\theta}-1}{e^{2i\theta}-1}=\frac{i-1}{e^{2i\theta}-1}=\frac{(i-1)e^{-i\theta}}{e^{i\theta}-e^{-i\theta}}=\frac{(i-1)(cos\theta-i\sin\theta)}{2i\sin\theta}.$$ Taking the real part, $$\sum_{r=0}^{n-1}\cos\frac{\pi r}{2n}=\frac{\cos\theta+\sin\theta}{2\sin\theta}=\frac12\cot\theta+\frac12.$$

In the limit when $\theta\rightarrow 0$, we can replace $\cot\theta$ by $\frac1{\theta}$ and ignore the term $\frac12$: $$\frac1n\sum_{r=0}^{n-1}\cos\frac{\pi r}{2n}\rightarrow\frac{1}{2n\theta}=\frac2{\pi}.$$

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By the Riemann sum we have

$$\lim_{n\to\infty}\frac1n\sum_{r=0}^{n-1}\cos\left(\frac{r\pi}{2n}\right)=\int_0^1\cos\left(\frac{\pi x}{2}\right)dx=\frac2\pi\sin\left(\frac{\pi x}{2}\right)\Bigg|_0^1=\frac2\pi$$

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this is nothing but $$\int_0^1dx \cos \frac{\pi x}{2}=\frac{2}{\pi}$$. This method is known as "limit of sum", where $\frac{r}{n} $ is replaced by $x$, $$\frac{1}{n}$$ is replaced by $dx$ and $$\sum_{n_0}^{n_1}$$ by $$\int_a^b$$. The limits are evaluated as: $$a=\lim_{n \to \infty}\frac{n_0}{n}$$ and $$b=\lim_{n \to \infty} \frac{n_1}{n}$$

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  • $\begingroup$ How did you convert it to integration ? $\endgroup$ – User Not Found Jun 25 '14 at 9:30

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