I know that the partial sums of $$\sum_{n=1}^{\infty}\sin(n)$$ are bounded between $\frac{\cos\left(\frac{1}{2}\right)-1}{2\sin\left(\frac{1}{2} \right)}$ and $\frac{1+\cos\left(\frac{1}{2} \right)}{2\sin\left(\frac{1}{2}\right)}$.

On the other hand, the partial sums of $$\sum_{n=1}^{\infty}\sin(\sqrt{n})$$ are unbounded.

I think that the partial sums of $\sum_{n=1}^{\infty}\sin(n^a)$ are bounded for $a \geq 1$ and unbounded for $0< a<1$, but how can I prove this? I think this question involves Euler-Maclaurin sum.

  • $\begingroup$ You mean the partial sums of these series are bounded/unbounded. $\endgroup$ – zhw. Nov 10 '18 at 20:12
  • 1
    $\begingroup$ It would be good to edit to reflect this. $\endgroup$ – zhw. Nov 10 '18 at 20:26
  • 1
    $\begingroup$ The partial sums of $\sum \sin(n^\alpha)$ are not bounded for $\alpha>1$, but $\sum_{n=1}^{N}\sin(n^\alpha)$ can be suitably controlled by a power of $N$ times a power of $\log N$. See Weyl's inequality, giving $$\sum_{n=1}^{N}\sin(n^2) \ll \sqrt{N}\log^2 N,$$ for instance. $\endgroup$ – Jack D'Aurizio Nov 10 '18 at 22:11
  • 1
    $\begingroup$ If you could achieve such bound $\sqrt{N}\log^2 N$, it solves your first question in math.stackexchange.com/questions/215528/…. When I tried it years ago, it did not work out very well, I could obtain $\alpha\leq 7/8$. If you managed to prove such bound, please include your method as an answer. $\endgroup$ – Sungjin Kim Nov 11 '18 at 3:51
  • 1
    $\begingroup$ @JackD'Aurizio You mentioned that the partial sums are unbounded when $a>1$. Would you please include that as an answer? I know Terry Tao solved for $a>1$ integer case in here mathoverflow.net/questions/201250/… $\endgroup$ – Sungjin Kim Nov 12 '18 at 18:14

$\bullet$ $a>1$ is an integer

If $a>1$ is an integer, then this post in MO by Terry Tao solves that the partial sums are not bounded.

$\bullet$ $a>1$ is not an integer

If $a>1$ is not an integer, the same argument by Terry Tao still works, since we have equidistribution of $n^a$ mod $2\pi$, and any sum or difference of $(n+i)^a$ with $1\leq i \leq h$ modulo $2\pi$. Let $X_i$ be the random variable $\sin^a (k+i)$ where $k=1,\ldots n$. Assuming the boundedness of partial sums, we end up having a contradiction that the random variables $X_1, \ldots, X_h$ such that $\mathrm{Var}(X_1+\cdots+X_h)$ is bounded as $h\rightarrow \infty$ by assumption, but $\mathrm{Var}(X_1+\cdots+X_h)\sim h/2$ as $h\rightarrow\infty$.

$\bullet$ $0<a<1$

The first part of this answer of mine, shows that $\sum_{\alpha<n\leq \beta} \sin(n^a)$ can be arbitrarily large. Thus, unboundedness of the partial sums follows.

For a better estimate, we apply Lemma 4.8 of The Theory of the Riemann Zeta-function written by Titchmarsh.

Let $f(x)$ be a real differentiable function in the interval $[a,b]$, let $f'(x)$ be monotonic, and let $|f'(x)|\leq \theta <1$. Then $$ \sum_{a<n\leq b}e^{2\pi i f(n)} = \int_a^b e^{2\pi i f(x)} \ dx + O(1). $$

Taking imaginary part from the lemma and $f(n)=n^a/(2\pi)$, we have

$$ \sum_{n\leq N} \sin(n^a) = \int_{1-}^N \sin(x^a) \ dx + O(1). $$

The change of variable $x^a=t$ gives $$ \int_{1-}^N \sin(x^a) \ dx=\int_{1-}^{N^a} \frac1a t^{\frac1a-1}\sin t \ dt. $$

Applying the integration by parts to the last integral, we obtain an estimate of $$-\frac1a N^{1-a}\cos(N^a) + O(N^{\max\{0,1-2a\}}).$$ This expression is clearly unbounded. Therefore, the partial sums are unbounded when $0<a<1$, and $a>1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.