# Is $\sum_{k=1}^{n} \sin(k^2)$ bounded by a constant $M$?

I know $\sum_{k=1}^{n} \sin(k)$ is bounded by a constant. How about $\sum_{k=1}^{n} \sin(k^2)$?

• @Elaqqad $M$ must be a constant. $n$ isn't a constant. – Scientifica Mar 27 '15 at 17:29
• Numerically, I find $\sum_{k=1}^{10^6}\sin(k^2)\approx -949.62$, so this particular sum is in the order of $\sqrt n$, which would be consistent with the summands being completely random ... But $n=10^7$ gives us $\approx 677.9$, so did we hit a ceiling? – Hagen von Eitzen Mar 27 '15 at 17:29
• 3 votes for closing this question (with 6 upvotes) reason: "off topic". I love MSE, but this is getting ridiculous. – leonbloy Mar 27 '15 at 18:37
• @EricNaslund : when a question is perceived (rightly or not) as too "imperative" , not showing effort or homework-like (the most common motive of close/downvote today, IMO) the user who wants to close it or tell the user to improve it, gets the frustrating-idiotic clasification system of MSE, where none of the given alternatives fits. So, there are tons of questions that are closed as "Off topic" when the problem is another. This is has been reported/discussed many times to meta, with no luck. – leonbloy Mar 27 '15 at 20:33
• @ leonbloy I agree. The term "off topic" here at MSE has always seemed to me woefully "off-the-mark". The question is clearly not off topic. It's an easily understood question (and an interesting one) about mathematics. If MSE wants a designation that signals there is deficiency in the question, according to criteria clearly stated somewhere at MSE, then fine. But don't call it "off topic" in a situation like this. – zhw. Feb 1 '17 at 1:29

We have: $$\begin{eqnarray*}2\left(\sum_{k=1}^{n}\sin(k^2)\right)^2 &=& \sum_{j,k=1}^{n}\cos(j^2-k^2)-\sum_{j,k=1}^{n}\cos(j^2+k^2)\\&=&n+2\sum_{m=1}^{n^2-1}d_1(m)\cos(m)-2\sum_{m=2}^{2n^2}d_2(m)\cos m\end{eqnarray*}$$ where $d_1(m)$ accounts for the number of ways to write $m$ as $j^2-k^2$ with $1\leq k<j\leq n$ and $d_2(m)$ accounts for the number of ways to write $m$ as $j^2+k^2$ with $1\leq j,k\leq n$. Since both these arithmetic functions do not deviate much from their average order (by Dirichlet's hyperbola method $d_1(m)$ behaves on average like $\log m$ and $d_2(m)$ behaves on average like $\frac{\pi}{4}$), I think it is not difficult to prove that for infinitely many $n$s $$\left|\sum_{k=1}^{n}\sin(k^2)\right|\geq C\sqrt{n}$$ holds for some absolute constant $C\approx \frac{1}{\sqrt{2}}$ through the Cauchy-Schwarz inequality.
However, this is quite delicate: Weyl bounds work just in the opposite direction. As an alternative approach, we may consider only the set of $n$ that appear in the numerators of the convergents of $2\pi$, in order that our sum behaves like a Gaussian sum (with magnitude $\sqrt{n}$ or $\sqrt{2n}$) plus a small error. In both cases we get that the sequence given by $$S_n=\sum_{k=1}^{n}\sin(k^2)$$ is not bounded.
• Thanks a lot. It's interesting that $T_n=\sum_{k=1}^{n}\sin(k)$ is bounded while $S_n$ is not. I'm just curious why the square term makes it so different. – Stan Mar 27 '15 at 19:10
• Also, it doesn't break equidistribution. $k^2$ is equidistributed modulo $2\pi$. This is one of the things that Weyl's inequalities prove. What's happening instead is that the rate of convergence is different. – Eric Naslund Mar 27 '15 at 20:39
• @npbool: the fact that $S_n\leq C\sqrt{n\log n}$ is well-known and follows from almost the same argument, but here we are interested in lower bounds, not in upper bounds. – Jack D'Aurizio Mar 28 '15 at 15:06
• How to prove that for infinitely many $n$s, $|S_n|\geq C\sqrt{n}$？Could you give me some hints? I find it difficult to estimate $d_1(m)$ and $d_2(m)$. – Xiang Yu Mar 30 '15 at 13:48