Calculate the limit: $\lim_{n\to+\infty}\sum_{k=1}^n\frac{\sin{\frac{k\pi}n}}{n}$ Using definite integral between the interval $[0,1]$.

Question

It seems to me like a Riemann integral definition:

$\sum_{k=1}^n\frac{\sin{\frac{k\pi}{n}}}{n}=\frac1n(\sin{\frac{\pi}n}+...+\sin\pi)$

So $\Delta x=\frac 1n$ and $f(x)=\sin(\frac{\pi x}n)$ (Not sure about $f(x))$

How do i proceed from this point?

You're absolutely on the right track. The point now is to realize this as a Riemann sum of what function? The limit should then be the value of that integral, which can be evaluated using the fundamental theorem of calculus. — Mark McClure, May 25 '13 at 14:57

score 6 · Answer 1 · edited Apr 13 '17 at 12:20

6

HINT:

Using this, $$\sum_{1\le r\le n}\sin \frac{\pi r}n=\frac{\sin \left( n\cdot \frac{\pi}{2n}\right)}{\sin \frac \pi{2n}}\cdot \sin \left(\frac{2\frac\pi n+(n-1)\frac{\pi}n}2\right)=\frac{\sin\left(\frac{(n+1)\pi}{2n}\right)}{\sin \frac \pi{2n}}$$

Now, $\sin\left(\frac{(n+1)\pi}{2n}\right)=\sin\left(\frac\pi2+\frac\pi{2n}\right)=\cos \frac\pi{2n}\implies \lim_{n\to\infty}\sin\left(\frac{(n+1)\pi}{2n}\right)=1$

$$\lim_{n\to\infty}\frac 1{n\sin \frac \pi{2n}}=\lim_{h\to0}\frac {2h}{\pi\sin h }(\text{ putting } \frac \pi{2n}=h)$$

$$\implies \lim_{n\to\infty}\frac 1{n\sin \frac \pi{2n}}=\frac2\pi\cdot\lim_{h\to0}\frac h{\sin h }=\frac2\pi$$

edited Apr 13 '17 at 12:20

Community

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answered May 25 '13 at 15:13

lab bhattacharjee

274,582

Thanks for the answer, but it didn't use the definite integral as asked. – StationaryTraveller May 25 '13 at 15:22
@TheAlchemist: And that makes it a special proof. – Inceptio May 25 '13 at 15:24
@TheAlchemist, do you know how to calculate definite integral from the definition , not as anti-derivative – lab bhattacharjee May 25 '13 at 15:49
Related : http://math.stackexchange.com/questions/305905/whats-the-justification-for-defining-the-definite-integral-using-a-countably – lab bhattacharjee May 25 '13 at 15:59

DonAntonio · Accepted Answer · 2013-05-25T15:52:29.880

4

Ideas: take the partition $\,P_n\,$ of the unit interval

$$P_n:=\left\{0=x_0\,,\,x_1=\frac1n\,,\ldots,\,x_k=\frac kn\,,\ldots,\,x_n=\frac nn=1\right\}$$

and choose points

$$c_i:=\frac in\;,\;\;1\le i\le n$$

Since we already know that $\,\sin \pi x\,$ is continuous everywhere and Riemann integrable in any closed, finite interval, we can apply the Riemann integral definition for the particular partitions and particular points $\,c_i\,$ as above in each subinterval, and get:

$$\lim_{n\to\infty}\frac 1n\sum_{k=1}^n \sin\frac{\pi k}n=\int\limits_0^1 \sin\pi x\,dx=\ldots\ldots$$

edited May 25 '13 at 15:52

answered May 25 '13 at 15:19

DonAntonio

211,718
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I think i got it, Thanks. – StationaryTraveller May 25 '13 at 15:24
Any time, @TheAlchemist – DonAntonio May 25 '13 at 15:26
1

@DonAntonio The clarity and brevity of your answer was very good. – Elias Costa May 25 '13 at 15:39

Elias Costa · Answer 3 · 2013-05-25T15:34:34.827

This is a problem that requires proper handling of the concept of Riemann sum and its relationship to definite integral. Consider the following partition of the interval $[0,1]$: $$ \mathscr{P}\Big( [0,1]\Big)=\left \{ 0=x_0<x_1=\frac{1}{n}<\ldots < x_k=k\cdot \frac{1}{n}<\ldots<x_{n}=n\cdot\frac{1}{n} \right\} $$ chose $x^*_k\in [x_{k-1},x_{k}]$ equal to $k\cdot \frac{1}{n}$ and set $\Delta x_k=x_k-x_{k-1}$ . Then we have $$ \sum_{k=1}^{n}\inf_{\xi\in[x_k,x_{k-1}]}\sin( \xi\pi)\cdot\Delta x_k \leq \sum_{k=1}^{n}\sin( x_k^*\pi)\cdot\Delta x_k \leq \sum_{k=1}^{n}\sup_{\xi\in[x_k,x_{k-1}]}\sin( \xi\pi)\cdot\Delta x_k $$ that is $$ \sum_{k=1}^{n}\inf_{\xi\in[x_k,x_{k-1}]}\sin( \xi\pi)\cdot\Delta x_k \leq \sum_{k=1}^n\sin{\left(\frac{k\pi}{n}\right )}\frac{1}{n} \leq \sum_{k=1}^{n}\sup_{\xi\in[x_k,x_{k-1}]}\sin( \xi \pi)\cdot\Delta x_k. $$ If we have $$ \lim_{n \to \infty}\sum_{k=1}^{n}\inf_{\xi\in[x_k,x_{k-1}]}\sin( \xi \pi)\cdot\Delta x_k = \int_0^1\sin( x\pi)dx = \lim_{n \to \infty}\sum_{k=1}^{n}\sup_{\xi\in[x_k,x_{k-1}]}\sin( \xi\pi)\cdot\Delta x_k $$ then \begin{align} \lim_{n\to+\infty}\sum_{k=1}^n\sin{\left(\frac{k\pi}{n}\right )}\frac{1}{n}= \int_0^1\sin( x\pi)dx \end{align}

Alex · Answer 4 · 2013-05-25T17:04:37.040

2

I think it should be $$ \int_{0}^{1}\sin \pi xdx=\frac{2}{\pi} $$

EDIT OK, the point here is the direct implementation of Riemanns sums with $d x = \frac{b-a}{n}, \ b=1, \ a=0$ and $\frac{k}{n}=x$

edited May 25 '13 at 17:04

answered May 25 '13 at 14:58

Alex

19,262

How do you know? I mean, How did you understand that is the function you need to find? – StationaryTraveller May 25 '13 at 14:59
set $x=\frac{k}{n}$ to get lower and upper bounds – Alex May 25 '13 at 15:00
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So, Shouldn't they be 0 and +1? And, By setting $x=\frac kn$ you mean i always take the right part of the squares? – StationaryTraveller May 25 '13 at 15:03
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Agree, $x=1$ if $k=n$. Sorry about the mistake. – Alex May 25 '13 at 15:18
I don't see the total purpose of down-voting? Rather than correcting just the $-1$ part, why the down vote? (+1) anyway. – Inceptio May 25 '13 at 15:21
1

Some people just can't wait to downvote instead of trying to give advice, to point the mistake, etc. Anyway, as this answer's lacking all explanation I can't upvote it, but I would never downvote it. – DonAntonio May 25 '13 at 15:25
OK I added an edit. Sorry about the mess. I do find other answere more substantial and to the point. – Alex May 25 '13 at 17:05

score 2 · Answer 5 · answered May 25 '13 at 15:28

Remember that if $f: [a,b] \to \mathbb{R}$ is some function we can define the Riemann Sum of $f$ in the interval by choosing a partition $P = \{t_0, \dots, t_n\}$ of $[a,b]$ in a very specific manner: we simply divide the interval $[a,b]$ equally calculating it's length $b-a$ and then dividing by the number of subintervals that we want. So a partition with $n$ subintervals will have $t_i = a + i\Delta t$ where $\Delta t = (b-a)/n$ is the size of each interval. In that case the integral of $f$ on the interval $[a,b]$ will be the limit:

$$\int_a^b f(t)dt=\lim_{n\to \infty} \sum_{i=1}^n f(t_i)\Delta t$$

Look that in your case we are working on the interval $[0,1]$ so for a partition of $n$ subintervals we'll have $\Delta t = 1/n$. Now we want to compute the limit of the following sum:

$$\sum_{i=1}^n \frac{1}{n}\sin{\left(\frac{i\pi}{n}\right)}$$

Note that with this we have:

$$\sum_{i=1}^n \Delta t\sin{\left(i\Delta t \pi\right)}$$

However, since the interval starts on $0$ and since $t_i = a + i\Delta t$, we have that $t_i = i\Delta t$, so we have:

$$\sum_{i=1}^n \sin{\left(t_i \pi\right)}\Delta t$$

Now on the limit this will become obviously the following integral:

$$\lim_{n\to \infty} \sum_{i=1}^n \sin{\left(t_i \pi\right)}\Delta t=\int_0^1 \sin(\pi t)dt$$

This integral is computed trivially by setting $u=\pi t$, $dt=du/\pi$ and so $u$ ranges from $0$ to $\pi$ and we have:

$$\lim_{n\to \infty} \sum_{i=1}^n \sin{\left(t_i \pi\right)}\Delta t=\frac{1}{\pi}\int_0^\pi \sin(u)du=\frac{2}{\pi}$$

Calculate the limit: $\lim_{n\to+\infty}\sum_{k=1}^n\frac{\sin{\frac{k\pi}n}}{n}$ Using definite integral between the interval $[0,1]$.

5 Answers5