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In Apostol Calculus I textbook (section 10.4, exercise 35e), he asks to prove the series sum:

$$ \lim_{n \rightarrow \infty}\sum_{k = 1}^n\frac{1}{n}\sin(\frac{k\pi}{n}) = \frac{2}{\pi}$$

I tried to do it by using the previous exercise: we assume that the above is the upper step function over the function $\sin(x)$. If the function is MONOTONIC on $[a, b]$, than we can use the identity:

$$ \lim_{n \rightarrow \infty}\sum_{k = 1}^n\frac{(b - a)}{n}f(a + \frac{(b-a)k}{n}) = \int_a^bf(x)dx$$

I specifically highlighted the word monotonic! The problem is that on the interval $0 - \pi$, sine will first increase, and then decrease, which renders the formula incorrect (according to the textbook previous exercise). WITHOUT the assumption about monotonicity, the same result as Apostol's is easily achieved. However, I doubt it after drawing the step function above the sine. I think the answer is incorrect. Also, I found some answers online which actually give the same answer as Apostol.

How can this be proven?

Bernard
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John
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  • You can split the sum in two parts which correspond to Riemann sums on $\Bigl[0,\frac \pi2\Bigr]$ and on $\Bigl[\frac \pi2,\pi\Bigr]$. – Bernard Nov 10 '19 at 22:56
  • @Bernard, yes, I tried to do that. However, I somehow got $4 / \pi$. – John Nov 10 '19 at 22:57
  • @John : Did you replace the upper limit of the sums with $n/2$ or change the widths of the rectangles to $1/(2n)$ when you split the sum? If not, then, yeah, you get two times the area. – Eric Towers Nov 10 '19 at 23:01
  • @EricTowers, I changed also the widths to $1/2n$. I got this function to integrate: $2/\pi \int_0^{\pi/2} sin(2x)dx$. However, it does not also help me, since I cannot get it: the $sin(2x)$ still changes from increasing to decreasing on this interval, so this whole conversion from series sum to integral is invalid (the identity i proved and wrote above was in previous exercise) – John Nov 10 '19 at 23:05
  • @EricTowers, oops i think i made a typo. I got $2/\pi$ after all. My bad. Though the question still stands ^ – John Nov 10 '19 at 23:08
  • Anyway, this sum is comprised between the upper and the lower Riemann sums, which both converge to the integral, so you have the result by the squeezing principle. – Bernard Nov 10 '19 at 23:13
  • @Bernard, yep, here I am confused. I did get this formula exactly by doing that. But it requires that the upper step function remains the upper step function on the whole interval of integration. Otherwise, we need to split the interval of integration. But the above series: suppose k is 1. Then we have $sin(\pi/n)$ - this is upper step function (from 0 to $1/n$). But if $k = (n + 1)/2$, $sin(\frac{(n+1)}{2}\pi/n)$ - it becomes the lower step function! It means we are summing up both upper and lower step function areas in the Riemann sum. – John Nov 10 '19 at 23:18
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    Not at all: on each small subinterval, there is an upper step and a lower step function,s, which are or not equal to $\sin\frac{ k\pi}n$, that is unimportant. What is important is that this term is between both. – Bernard Nov 10 '19 at 23:25
  • Ok, I need to think a little about all your info here :) Thx guys. – John Nov 10 '19 at 23:31
  • Your result regarding integrals holds for any Riemann integrable function by definition of Riemann integral. Why do you restrict it to only monotone functions and unnecessarily complicate your problem? – Paramanand Singh Nov 11 '19 at 02:13

2 Answers2

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The sine function is monotonic on $[0,\pi/2]$ and $[\pi/2, \pi]$. In the summation, this is the requirement $k < N/2$ and $k > N/2$, respectively. One way to go with this (the most directly, but perhaps not the clearest) is \begin{align*} \lim_{N \rightarrow \infty} &\sum_{k = 1}^N \frac{1}{N} \sin\left( \frac{k \pi}{N} \right) \\ &= \begin{cases} \lim_{N \rightarrow \infty} \sum_{k = 1}^N \frac{1}{N} \sin\left( \frac{k \pi}{N} \right) &, \text{$N$ odd} \\ \lim_{N \rightarrow \infty} \sum_{k = 1}^N \frac{1}{N} \sin\left( \frac{k \pi}{N} \right) &, \text{$N$ even} \end{cases} \\ &= \begin{cases} \lim_{N \rightarrow \infty} \left( \sum_{k = 1}^{\lfloor N/2 \rfloor} \frac{1}{N} \sin\left( \frac{k \pi}{N} \right) + \sum_{k = \lceil N/2 \rceil}^{N} \frac{1}{N} \sin\left( \frac{k \pi}{N} \right) \right) &, \text{$N$ odd} \\ \lim_{N \rightarrow \infty} \left( \sum_{k = 1}^{N/2} \frac{1}{N} \sin\left( \frac{k \pi}{N} \right) + \sum_{k = 1 + N/2}^{N} \frac{1}{N} \sin\left( \frac{k \pi}{N} \right) \right) &, \text{$N$ even} \end{cases} \\ &= \int_0^{1/2} \sin(\pi x) \,\mathrm{d}x + \int_{1/2}^{1} \sin(\pi x) \,\mathrm{d}x \\ &= \frac{1}{\pi} + \frac{1}{\pi} \\ &= \frac{2}{\pi} \text{.} \end{align*}

Here, sine is monotonic in each sum, so we can use the previous exercise, which you seem to want to do.

Eric Towers
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  • Thanks, Eric. This is exactly how I wanted to do it, since it addresses the problem I was referring to. I need to think about the points of others though. – John Nov 10 '19 at 23:31
  • Thank you, Eric. How do you know that you can split the original sum into 2 parts, and then that the convergence of the individual parts can be claimed based on the theorem the question mentions (please see https://math.stackexchange.com/a/4544937/942378 for context, if needed)? In the question (i.e. in the result we can use), $N$ in the limit and the sum length are the same, and in your case the $N$ in the limit is 2 times larger than the sum length, so can you please expand the answer to justify that part? – S11n Oct 05 '22 at 09:57
  • @EricTowers I expanded on your answer here: https://math.stackexchange.com/a/4546431/942378. – S11n Oct 06 '22 at 10:12
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    @S11n : We don't rearrange an infinite sum. Let $(N_i){i=1}^\infty$ be a sequence of positive integers with $\lim{i \rightarrow \infty} N_i = \infty$. Then we are studying $\lim_{i \rightarrow \infty} \sum_{k=1}^{N_i} \dots$. For each $N_i$, this is a finite sum and rearrangement requires no justification. If one still requires justification : the sine function is nonnegative on $[0,2\pi]$, so the limit series is either absolutely convergent or at least one of the two rearranged series diverges. We find that neither diverges so rearrangement back into the original series converges. – Eric Towers Oct 06 '22 at 12:27
  • @EricTowers Thanks! By rearrangement, did you mean splitting the original sum into two? My main question was not why you can split the sum into two, but why you could apply the result for $\sum_{k = 1}^{N} \frac{1}{2N} \sin \frac{k \pi}{2N}$ to $\sum_{k = 1}^{N} \frac{1}{2N+1} \sin \frac{k \pi}{2N+1}$. In my question I answered that in quite a verbose way, but I'm sure you have a more concise justification which I would prefer to see: https://math.stackexchange.com/a/4546431/942378. – S11n Oct 06 '22 at 17:31
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$$\lim_{n \rightarrow \infty}\sum_{k = 1}^n\frac{1}{n}\sin(\frac{k\pi}{n}) = \int _0^1 \sin (\pi x) dx = \frac{2}{\pi}$$

Bernard
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Mohammad Riazi-Kermani
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  • The $sin(\pi x)$ is not monotonically increasing on $[0, 1]$. Why is it allowed to do that? – John Nov 10 '19 at 22:55
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    It's just the limit of a Riemann sum. What does monotonicity have to do with it? – saulspatz Nov 10 '19 at 23:04
  • @saulspatz, because at $\pi/2$, the upper step function becomes the lower step function, since sine starts decreasing at that point. For this reason, this substitution above ^ is allowed only when the function is either increasing or decreasing on the interval. Well, at least this is what i proved in previous exercise. Am I wrong? – John Nov 10 '19 at 23:10
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    @John you can evaluate the function at any point in the subinterval. You don't have to take the upper or lower sum. Mohammad's answer is correct. Of course, I don't know exactly what definition of the integral you've seen up to this point, but if the upper and lower sum converge to the same value $I$, if you choose values between the max and min, the sum is squeezed between the upper and lower sums, and must converge to $I$ also. – saulspatz Nov 10 '19 at 23:20