# Boundedness of sum of sin(sin(n))

Playing with desmos I have accidentally noticed that the sequence of partial sums $$\left\{ \sum_{n=1}^{N}\sin(\sin(n)) : N\geq 1 \right\}$$

is bounded.

However, I did not succeed in proving this statement.

My main idea was to show that there exists some constant $$C > 0$$ such that

$$\forall N \geq 1 \; \forall k \geq 0 \; : \left\vert \sum_{n=1}^{N} (\sin(n))^{2k+1}\right\vert \leq C$$

And then using the Taylor series

$$\sin(\sin(n)) = \sum_{m=0}^{\infty}\frac{(-1)^m}{(2m+1)!}(\sin(n))^{2m+1}$$

bound the partial sums defined above by $$Ce$$.

Could you please help me or suggest the techniques which may be useful in proving statements of this kind.

• As to your approach, I think even the partial sums of $\sin(n)$ are $\underline{\text{un}}$bounded. Commented Aug 5 at 11:31
• @mathworker21 I don't think that's true: Writing $\sin(n)$ as imaginary part of $e^{in}$, the partial sums evaluate to the imaginary part of a geometric sum $e^i \cdot \frac{e^{iN}-1}{e^i-1}$, where the numerator stays bounded by $2$ no matter how large $N$ gets. Commented Aug 5 at 11:46
• @AchimKrause Thanks. Don't know why I thought otherwise. Commented Aug 5 at 12:34

This is true. It doesn't have much to do with the details of $$\sin(\sin(\ ))$$. Rather:

Theorem Let $$f: \mathbb{R} \to \mathbb{R}$$ be any smooth, $$2 \pi$$-periodic function with $$\int_{z=0}^{2 \pi} f(z) dz=0$$. Then $$\sum_{n=1}^N f(n)$$ is bounded independent of $$N$$.

There are two ingredients in the proof. First, we bound the coefficients of the Fourier series of $$f$$. We can write $$f(\theta) = \sum_{k=-\infty}^{\infty} a_k e^{i k \theta}$$ where $$a_k = \frac{1}{2 \pi} \int_{\theta=0}^{2 \pi} e^{-i k \theta} f(\theta).$$ We have $$a_0=0$$ by the assumption on the integral of $$f$$. A standard strengthening of the Riemann-Lebesgue lemma says that, if $$f$$ is $$C^r$$, then $$a_k = o(1/k^r)$$. (See, for example, here.) In our case, $$f$$ is smooth, so we obtain that $$a_k = O(1/k^r) \ \text{for every}\ r. \tag{1}$$

So $$\sum_{n=1}^N f(n) = \sum_{k \neq 0} a_k \sum_{n=1}^N e^{ikn} = \sum_{k \neq 0} a_k \frac{e^{ik(N+1)} - e^{ik}}{e^{ik}-1} \tag{2}$$ where there is no $$k=0$$ term since $$a_0=0$$.

Now, Zeilberger and Zudlin proved that, for any integers $$k$$ and $$\ell \neq 0$$, we have $$|\pi - k/\ell| \geq c/\ell^{7.11}$$ for some $$c>0$$. So $$|k - \ell \pi| \geq c/\ell^{6.11} \geq c'/k^{6.11}$$ for some other constant $$c'$$. So $$\frac{e^{ik(N+1)} - e^{ik}}{e^{ik}-1} = O(k^{6.11}). \tag{3}$$

Combining $$(1)$$ and $$(3)$$, the summand in (2) is $$O(k^{6.11-r})$$ for every $$r$$, and thus in particular is $$O(k^{-2})$$. So the sum in $$(2)$$ is bounded, independent of $$N$$.

• This is a great answer. It shows how the ultimate point is that $\pi$ is not well approximated by rational numbers. Commented Aug 5 at 12:06
• Thanks! I've been wondering whether I can find a $\theta$ for which the sums of $\sin(\sin(n \theta))$ are unbounded. The tricky thing is that, if $\sin(\sin(\phi))$ is close to its maximum, then $\sin(\sin(3 \phi))$ is close to its minimum, so you tend to get a lot of cancellation. Commented Aug 5 at 12:19
• Mr. David, thank you very much for your solution, the theorems you stated are really nice and useful ! I did not know about them before. Commented Aug 5 at 13:08
• If I understand correctly, the only information needed about $\pi$ is that its irrationality measure is finite.
– abx
Commented Aug 5 at 14:23
• Seems that ZZ's bound is overkill for this theorem. How cheaper is it to prove a coarser bound, good enough to do the job? (Differently put: do I have a reasonable chance to make a problem out of this wonderful answer?) Commented Aug 5 at 15:54

I think one can use similar tricks as in the accepted answer at Is the series $\sum_n|\sin n|^n/n$ convergent?.

To start with, I don't think one can hope for a constant bound $$C$$ as you hoped. Indeed,

$$\sum_{n=1}^N \sin(n)^{2k+1} = \frac{1}{(2i)^{2k+1}} \sum_{n=1}^N (e^{ni}-e^{-ni})^{2k+1}$$ $$= \frac{1}{(2i)^{2k+1}} \sum_{n=1}^N \sum_{l=0}^{2k+1} \binom{2k+1}{l} e^{(2k+1-2l)in}$$ $$= \frac{1}{(2i)^{2k+1}} \sum_{l=0}^{2k+1} \binom{2k+1}{l} \sum_{n=1}^N e^{(2k+1-2l)in}$$ $$= \frac{1}{(2i)^{2k+1}} \sum_{l=0}^{2k+1} \binom{2k+1}{l} e^{(2k+1-2l)i}\frac{e^{(2k+1-2l)iN}-1}{e^{(2k+1-2l)i}-1}.$$ This stays bounded for large $$N$$, as the numerators are bounded by $$2$$. However, the denominator here is controlled by how close $$2k+1-2l$$ gets to a multiple of $$2\pi$$, and so is related to rational approximations of $$\pi$$. So I believe these sums will not admit a uniform constant bound. But as in the linked answer, there is some lower bound on the quality of rational approximations of $$\pi$$. Let us write $$d_k = \min_{0\leq l\leq 2k+1, q\in \mathbb{Z}} |2k+1-2l-2q\pi|.$$ Thus, $$d_k = 2q \cdot |\frac{2p+1}{2q}-\pi|$$ where $$\frac{2p+1}{2q}$$ is the best approximation to $$\pi$$ with odd numerator $$|2p+1|\leq 2k+1$$ and even denominator. By the linked answer, we have $$d_k > 2q\frac{1}{(2q)^{\mu+\varepsilon}} > \frac{1}{(2q)^{\mu+\varepsilon-1}} > \frac{1}{(2k+1)^7}.$$

Now $$|e^{(2k+1-2l)-1}-1| > C\cdot d_k$$ for some global constant $$C$$ and all $$0\leq l\leq 2k+1$$, and so we obtain a bound

$$\left| \sum_{n=1}^N \sin(n)^{2k+1} \right| < \frac{2}{C d_k} < \frac{1}{C} (2k+1)^7.$$

This yields a uniform upper bound for $$|\sum_{n=1}^N \sin(\sin(n))|$$ of the form $$\frac{1}{C} \sum_{k=0}^\infty \frac{(2k+1)^7}{(2k+1)!},$$ which is finite.

• Dear Achim Krause, I am grateful for you solution. It is also great ! Thank you for teaching me new tricks :) Commented Aug 5 at 13:10