Question

What is $\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left( {\sin \left( {\dfrac{\pi }{n}} \right) + \sin \left( {\dfrac{{2\pi }}{n}} \right) + .......... + \sin \left( {\dfrac{{2n\pi }}{n}} \right)} \right)$?

Answer 1

Here, we will first write the general term for the expression given in the question. Then, we will use the concepts of the limit of sum to change the expression into an integration form. Then, we will write the limits of integration and then will carry out the integration.
Formula used:
$\cos 2\pi = 1$
$\cos 0 = 1$
$\int {\sin \left( {nx} \right)dx = \dfrac{{ - \cos \left( {nx} \right)}}{n}}$

Complete step by step answer:
Given,
$\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left( {\sin \left( {\dfrac{\pi }{n}} \right) + \sin \left( {\dfrac{{2\pi }}{n}} \right) + .......... + \sin \left( {\dfrac{{2n\pi }}{n}} \right)} \right)$
Now, we will first write the general term for the expression given in the question. We can observe that:
1st term is: $\sin \dfrac{\pi }{n}$
2nd term is: $\sin \dfrac{{2\pi }}{n}$
Last term is: $\sin \dfrac{{2n\pi }}{n}$
So, we can say that the angle $\dfrac{\pi }{n}$ is constant and its coefficient is increasing from $1{\text{ to }}2n$.
Hence, we can write the general term as:
$\sum\limits_{r = 1}^{2n} {\sin \left( {\dfrac{{r\pi }}{n}} \right)}$
Now, the question becomes:
$\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\sum\limits_{r = 1}^{2n} {\sin \left( {\dfrac{{r\pi }}{n}} \right)}$
Now, for evaluating the above expression we will use the concepts of limit of sum.
So,
$\dfrac{1}{n} \to dx$, and $\dfrac{r}{n} \to x$.
Now, on the summation sign we can see that $r{\text{ goes from }}1{\text{ to }}2n$. From here we will decide the limit of integration.
Since, $\dfrac{r}{n} \to x$, so we can write at:
$r = 1{\text{ we get }}x = \dfrac{1}{n}$
$\Rightarrow x = 0{\text{ since }}n \to \infty$
Similarly, at $r = 2n$ we get $x = \dfrac{{2n}}{n}$
$\Rightarrow x = 2$
Now, we will put these into $\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\sum\limits_{r = 1}^{2n} {\sin \left( {\dfrac{{r\pi }}{n}} \right)}$. So, we get
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\sum\limits_{r = 1}^{2n} {\sin \left( {\dfrac{{r\pi }}{n}} \right)} = \int\limits_0^2 {\sin \left( {\pi x} \right)} dx$
On integrating the RHS, the above equation becomes
$= \left( {\dfrac{{ - \cos \left( {\pi x} \right)}}{x}} \right)_0^2$
Now we will put the upper and lower limit.
$= \left( {\dfrac{{ - \cos \left( {2\pi } \right)}}{\pi } - \dfrac{{ - \cos \left( {0\pi } \right)}}{\pi }} \right)$
Now we will put $\cos 2\pi = 1$, and $\cos 0 = 1$, we get
$= \left( {\dfrac{{ - 1}}{\pi } + \dfrac{1}{\pi }} \right)$
$= 0$
So,
$\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left( {\sin \left( {\dfrac{\pi }{n}} \right) + \sin \left( {\dfrac{{2\pi }}{n}} \right) + .......... + \sin \left( {\dfrac{{2n\pi }}{n}} \right)} \right)$ $= 0$

Note:
One major mistake that most of the students commit is that they do not change the limit when they convert the expression to the integration form. So, take care that while changing to integration form change the limit of summation accordingly. Another point to note is that in the summation the limit of $r$ goes from $1{\text{ to }}2n$ and not from ${\text{0 to }}2n$.