In the solution @Jack D'Aurizio: gave the estimate $$\sum_{k=1}^{n} \cos k^2 = \mathcal{O}(\sqrt{n}\log n)$$ and mentioned the Weyl's inequalities
$$\sum_{k=1}^n \cos( f(k) ) = \mathcal{O}(F(n))$$ if $f$ is polynomial function.

I wonder if one can give some nontrivial estimates for other kind of functions, for instance $$\sum_{k=1}^n \cos \sqrt{k} = \mathcal{O}(F(n))$$ for some $F(n) = \mathcal{o}(n)$.

  • $\begingroup$ @LeGrandDODOM: your statement is interesting... I wonder why that is true. $\endgroup$ – Orest Bucicovschi Aug 18 '17 at 13:41

We have

$$\sum_{k = 1}^n \cos \sqrt{k} = \frac{\cos 1 + \cos \sqrt{n}}{2} + \int_1^n \cos \sqrt{t}\,dt - \int_1^n \bigl(\lbrace t\rbrace - \tfrac{1}{2}\bigr)\frac{\sin \sqrt{t}}{2\sqrt{t}}\,dt.$$

The first term is bounded, and the second integral is $O(\sqrt{n})$ as one sees from the boundedness of $\sin \sqrt{t}$ and $\lbrace t\rbrace - \frac{1}{2}$. It thus remains to estimate

\begin{align} \int_1^n \cos \sqrt{t}\,dt &= 2\int_1^{\sqrt{n}} u\cos u\,du \\ &= 2u\sin u \biggr\rvert_1^{\sqrt{n}} - 2\int_1^{\sqrt{n}} \sin u\,du \\ &= 2\sqrt{n}\sin \sqrt{n} - 2\sin 1 + 2\cos \sqrt{n} - 2, \end{align}

which clearly is in $O(\sqrt{n})$.

Generally, for $0 < \alpha < 1$, we obtain

$$\sum_{k = 1}^n \cos k^{\alpha} = \frac{\cos 1 + \cos n^{\alpha}}{2} + \int_1^n \cos t^{\alpha}\,dt - \alpha\int_1^n p_1(t)t^{\alpha-1}\sin t^{\alpha}\,dt,\tag{$\ast$}$$

where $p_1(t) = \lbrace t\rbrace - \frac{1}{2}$. We can estimate the first integral substituting $u = t^{\alpha}$ and integrating by parts:

\begin{align} \int_1^n \cos t^{\alpha}\,dt &= \frac{1}{\alpha} \int_1^{n^{\alpha}} u^{\frac{1}{\alpha}-1}\cos u\,du \\ &= \frac{u^{1/\alpha-1}\sin u}{\alpha}\biggr\rvert_1^{n^{\alpha}} - \frac{1}{\alpha}\biggl(\frac{1}{\alpha}-1\biggr)\int_1^{n^{\alpha}} u^{\frac{1}{\alpha}-2}\sin u\,du. \end{align}

The first term is $\frac{1}{\alpha}\bigl(n^{1-\alpha}\sin n^{\alpha} - \sin 1\bigr)$, and using $\lvert \sin u\rvert \leqslant 1$ we see that the remaining integral also belongs to $O(n^{1-\alpha})$.

For the integral

$$\alpha\int_1^n p_1(t)t^{\alpha-1}\sin t^{\alpha}\,dt,$$

we immediately obtain an $O(n^{\alpha})$ bound using the boundedness of $p_1$ and $\sin$. For $\alpha \leqslant \frac{1}{2}$, this is smaller than the bound on the other integral. For $\alpha > \frac{1}{2}$, we still have an $O(n^{\alpha})$ bound for the sum, which suffices to conclude that

$$\sum_{n = 1}^{\infty} \frac{\cos k^{\alpha}}{k}$$

converges. But we can lower the bound using integration by parts:

$$\int_1^n p_1(t) t^{\alpha-1} \sin t^{\alpha} = p_2(t)t^{\alpha-1}\sin t^{\alpha}\biggr\rvert_1^n - (\alpha-1)\int_1^n p_2(t)t^{\alpha-2}\sin t^{\alpha}\,dt - \alpha \int_1^n p_2(t)t^{2(\alpha-1)}\cos t^{\alpha}\,dt$$

where the first two terms on the right are bounded, and the last integral is elementarily bounded by $C\cdot n^{2\alpha - 1}$. Continuing integration by parts, we find that the last integral in $(\ast)$ belongs to $O(n^{1 - m(1-\alpha)})$ for every $0 < m \leqslant \frac{1}{1-\alpha}$, and then eventually we obtain

$$\int_1^n p_1(t)t^{\alpha-1}\sin t^{\alpha}\,dt \in O(1),$$

so the sum belongs to $O(n^{1-\alpha})$ for every $\alpha \in (0,1)$ [this is also true for $\alpha = 0$ and $\alpha = 1$].

Noting that $\cos k^{\alpha} \geqslant \frac{1}{2}$ for

$$\bigl(2m - \tfrac{1}{3}\bigr)\pi \leqslant k^{\alpha} \leqslant \bigl(2m + \tfrac{1}{3}\bigr)\pi,$$

we see that the bound above is sharp. For $n \approx \bigl((2m+\frac{1}{3})\pi\bigr)^{1/\alpha}$, we have $\Theta(n^{1-\alpha})$ successive terms that are $\geqslant \frac{1}{2}$, whence the partial sum must have had at least order $\ell^{1-\alpha}$ for some $\ell \leqslant n$.

  • $\begingroup$ Euler summation, great! $\endgroup$ – Orest Bucicovschi Aug 18 '17 at 13:09
  • $\begingroup$ Could you get in a similar way estimates for other functions, say $f(n) = n^{\frac{3}{4}}$ ? It's just that the second integral in this case does not appear that nice. $\endgroup$ – Orest Bucicovschi Aug 18 '17 at 13:20
  • $\begingroup$ Yes, we can get estimates for $\sum \cos k^{\alpha}$, $0 < \alpha < 1$. I'll take out some paper, and afterwards edit the results in. $\endgroup$ – Daniel Fischer Aug 18 '17 at 13:26
  • $\begingroup$ @orangeskid if $\sum_{k=1}^n (-1)^kk^{\frac 1{\alpha}-1}= O(g(n))$, then $\sum_{k=1}^n \cos k^{\alpha} = O(g(n^\alpha))$. But I don't know $g$... $\endgroup$ – Gabriel Romon Aug 18 '17 at 13:34
  • $\begingroup$ With Mathematica I get that the integrals $\int \cos(x^{\frac{1}{n}}) dx$ are elementary, $ \int \cos (x^{\frac{2}{n}}) dx$ involve Fresnel integral functions, and in general we get some incomplete Gamma's. Even in the last case there are some asymptotics. $\endgroup$ – Orest Bucicovschi Aug 18 '17 at 13:40

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.