Boundedness of sum of sin(sin(n))

Question

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.

Answer 1

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 Zudilin proved (doi link, arXiv) 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 $N$.

Answer 2

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.