Is $\sum_{k=1}^n \cos \sqrt k= o(n)$ as $n\to \infty$?

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 $F(n) = \mathcal{o}(n)$.

• @LeGrandDODOM: your statement is interesting... I wonder why that is true. – 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 $\ell \leqslant n$.

• Euler summation, great! – Orest Bucicovschi Aug 18 '17 at 13:09
• 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. – Orest Bucicovschi Aug 18 '17 at 13:20
• 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. – Daniel Fischer Aug 18 '17 at 13:26
• @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 $g$... – Gabriel Romon Aug 18 '17 at 13:34
• 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. – Orest Bucicovschi Aug 18 '17 at 13:40