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From exercise $10.4.34$ in Tom Apostol's Calculus vol $1$ book, we know that:

$\tag{1} \lim_{N \rightarrow \infty} \sum_{k = 1}^N \frac{1}{2N} \sin \frac{k \pi}{2N} = \int_0^{\frac{1}{2}}\sin(x \pi) dx $

And the task in $10.4.35.f$ is to find $\tag{2} \lim_{N \rightarrow \infty} \sum_{k = 1}^N \frac{1}{N} \sin \frac{k \pi}{N} $

Because $\sin$ function is not monotonic on the region $[0, \pi]$, I cannot apply $(1)$ directly (because the theorem we proved in $10.4.34$ assumed that the function is monotonic). Therefore, I tried splitting the domain into two $[0, \pi/2)$, $[\pi/2, \pi)$, but I was going in circles, because I would again not get the sum which I was supposed to solve. Then I found this answer (link), which looks quite close to what I wanted to do:

" \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*} "

However, the author there seems to take the following implication for granted, and I'm not sure why is that justified:

$\lim_{N \rightarrow \infty} \sum_{k = 1}^N \frac{1}{2N} \sin \frac{k \pi}{2N} = \int_0^{\frac{1}{2}}\sin(x \pi) dx \implies \lim_{N \rightarrow \infty} \sum_{k = 1}^{\lfloor N/2 \rfloor} \frac{1}{N} \sin( \frac{k \pi}{N}) = \int_0^{\frac{1}{2}}\sin(x \pi) dx $

Especially because the implication $\lim_{n \to +\infty} s_{2n} = L \implies \lim_{n \to +\infty} s_n = L$ does not hold in general (link).

Edit 1.

I will add one step how I could try to prove it using the definition of sequence convergence:

The definition of sequence conversion is a usual, where $m$ are positive integers:

$\lim_{m \to +\infty} s_m = L \iff [\forall \epsilon > 0. \exists M > 0. m > M \implies |s_m - L| < \epsilon]$

We know that: $\tag{3} \lim_{N \rightarrow \infty} \sum_{k = 1}^N \frac{1}{2N} \sin \frac{k \pi}{2N} = \frac{1}{\pi} \implies \\ [\forall \epsilon > 0. \exists M > 0. N > M \implies |\sum_{k = 1}^N \frac{1}{2N} \sin \frac{k \pi}{2N} - \frac{1}{\pi}| < \epsilon] \land \\ |\sum_{k = 1}^{N+1} \frac{1}{2N+2} \sin \frac{k \pi}{2N+2} - \frac{1}{\pi}| < \epsilon] \implies \\ \lim_{N \rightarrow \infty} \sum_{k = 1}^N \frac{1}{2N+2} \sin \frac{k \pi}{2N+2} = \frac{1}{\pi}$

If we take the odd case in the answer above, which I'm zooming in above, we want to prove that:

$\tag{4} \lim_{n \rightarrow \infty} \sum_{k = 1}^{\lfloor n/2 \rfloor} \frac{1}{n} \sin( \frac{k \pi}{n}) = \frac{1}{\pi}$

$(4)$ means

$\forall \epsilon' > 0. \exists M' > 0. n > M' \implies |\sum_{k = 1}^{\lfloor n/2 \rfloor} \frac{1}{n} \sin( \frac{k \pi}{n}) - \frac{1}{\pi}| < \epsilon'$

For odd $n$, we can let $n = 2N + 1$, so we get

$\forall \epsilon' > 0. \exists M' > 0. n = 2N + 1 > M' \implies |\sum_{k = 1}^{N} \frac{1}{2N + 1} \sin( \frac{k \pi}{2N + 1}) - \frac{1}{\pi}| < \epsilon'$

Which is different than the limit in $(3)$...

S11n
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  • What's the context of Apostol's course so far? To me, this should be immediate as a Riemann sum – FShrike Oct 05 '22 at 14:33
  • @FShrike Riemann sum was still not mentioned. You could see one proposed solutions in https://www.stumblingrobot.com/2016/02/29/establish-the-given-limit-relations/#comment-2646. It is trying to stick to the context provided in the book so far, but the solver there ignored the condition of function monotonicity, so it's not complete (in a sense of not using non-proven assumptions or non-stated axioms). – S11n Oct 05 '22 at 15:01
  • @FShrike Added more context to the question here... – S11n Oct 06 '22 at 08:58
  • @FShrike And added an answer... – S11n Oct 06 '22 at 10:12

1 Answers1

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We also know (link) that

$\tag{5} \lim_{N \rightarrow \infty} \sum_{k = 0}^{N-1} \frac{1}{2N} \sin \frac{k \pi}{2N} = \frac{1}{\pi} \implies \\ [\forall \epsilon > 0. \exists M > 0. N > M \implies |\sum_{k = 0}^{N-1} \frac{1}{2N} \sin \frac{k \pi}{2N} - \frac{1}{\pi}| < \epsilon \land \\ |\sum_{k = 0}^{N} \frac{1}{2N+2} \sin \frac{k \pi}{2N+2} - \frac{1}{\pi}| < \epsilon] \implies \\ [\forall \epsilon > 0. \exists M > 0. N > M \implies |\sum_{k = 1}^{N} \frac{1}{2N+2} \sin \frac{k \pi}{2N+2} - \frac{1}{\pi}| < \epsilon] \implies \\ \lim_{N \rightarrow \infty} \sum_{k = 1}^{N} \frac{1}{2N+2} \sin \frac{k \pi}{2N+2} = \frac{1}{\pi} $

We have:

$\sum_{k = 1}^{N} \frac{1}{2N+2} \sin \frac{k \pi}{2N+2} \le \sum_{k = 1}^{N} \frac{1}{2N+1} \sin \frac{k \pi}{2N+1} \le \sum_{k = 1}^{N} \frac{1}{2N} \sin \frac{k \pi}{2N}$

From $(3)$ and $(5)$, and by squeezing principle (Theorem 3.3. in the book), we have:

$\lim_{N \to +\infty} \sum_{k = 1}^{N} \frac{1}{2N+1} \sin \frac{k \pi}{2N+1} = \frac{1}{\pi}$

I answered my question, because I at least now have a solution, but please feel free to provide a simpler one, and I'm happy to accept it (if it's complete).

S11n
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