# Two questions about weakly convergent series related to $\sin(n^2)$ and Weyl's inequality

By using partial summation and Weyl's inequality, it is not hard to show that the series $\sum_{n\geq 1}\frac{\sin(n^2)}{n}$ is convergent.

• Is is true that $$\frac{1}{2}=\inf\left\{\alpha\in\mathbb{R}^+:\sum_{n\geq 1}\frac{\sin(n^2)}{n^\alpha}\mbox{ is convergent}\right\}?$$
• In the case of a positive answer to the previous question, what is $$\inf\left\{\beta\in\mathbb{R}^+:\sum_{n\geq 1}\frac{\sin(n^2)}{\sqrt{n}(\log n)^\beta}\mbox{ is convergent}\right\}?$$
• By modelling $\sin(n^2)$ as a sequence of independent random variables $X_n$, I would expect a positive answer to the first question, also I would expect the later series to be convergent when $\beta > 1/2$ and divergent when $\beta < 1/2$. Thus the answer to the second question would be $1/2$. – blabler Nov 30 '12 at 4:13
• A related question has popped up since this one was asked. – Douglas B. Staple Mar 29 '13 at 3:44
• Applying the argument here, math.stackexchange.com/questions/2270/… we obtain that the first quantity is $\leq \frac{7}{8}$ – Sungjin Kim Jun 8 '13 at 17:33

I recall a generalization of partial summation formula:

Suppose that $$\lambda_1,\lambda_2,\ldots$$ is a nondecreasing sequence of real numbers with limit infinity, that $$c_1,c_2,\ldots$$ is an arbitrary sequence of real or complex numbers, and that $$f(x)$$ has a continuous derivative for $$x\geq \lambda_1$$. Put $$C(x)=\sum_{\lambda_n\leq x}c_n,$$ where the summation is over all $$n$$ for which $$\lambda_n\leq x$$. Then for $$x\geq\lambda_1$$, $$\sum_{\lambda_n\leq x}c_nf(\lambda_n)=C(x)f(x)-\int^{x}_{\lambda_1}C(t)f'(t)dt.\tag 1$$

Now we can write if $$y=x^2$$ and $$\lambda_n=n^2$$ and $$C(t)=[\sqrt{t}]$$ (integer part of $$\sqrt{t}$$): $$S=\sum_{1\leq n\leq x}\frac{\sin(n^2)}{n^a}=\sum_{\lambda_n\leq y}\frac{\sin(\lambda_n)}{\lambda_n^{a/2}}=$$ $$=[\sqrt{y}]\frac{\sin(y)}{y^{a/2}}-\int^{y}_{1}[\sqrt{t}]\frac{d}{dt}\left(\frac{\sin(t)}{t^{a/2}}\right)dt.$$ But it is $$[\sqrt{t}]=\sqrt{t}-\{\sqrt{t}\}$$, where $$\{\sqrt{t}\}$$ is the fractional part of $$\sqrt{t}$$. Hence $$S=-\frac{1}{2}Re\left[iy^{1/2-a/2}E\left(\frac{1+a}{2},iy\right)\right]+\frac{1}{2}Re\left[iE\left(\frac{1+a}{2},i\right)\right]+\sin(1)-\{\sqrt{y}\}\frac{\sin(y)}{y^{a/2}}+$$ $$+\int^{y}_{1}\{\sqrt{t}\}\frac{d}{dt}\left(\frac{\sin(t)}{t^{a/2}}\right)dt,$$ where $$E(a,z)=\int^{\infty}_{1}\frac{e^{-tz}}{t^a}dt$$ But when $$a>0$$ and $$y\rightarrow+\infty$$ we have $$\lim_{y\rightarrow+\infty}\left\{-\frac{1}{2}Re\left[iy^{1/2-a/2}E\left(\frac{1+a}{2},iy\right)\right]+\frac{1}{2}Re\left[iE\left(\frac{1+a}{2},i\right)\right]\right\}+\sin(1)=$$ $$=\frac{1}{2}Re\left[iE\left(\frac{1+a}{2},i\right)\right]+\sin(1)$$ Also $$x$$ is positive integer and $$\{\sqrt{y}\}=0$$.

Hence when $$a>0$$, then $$\lim_{x\rightarrow\infty}\sum^{x}_{n=1}\frac{\sin(n^2)}{n^a}=\frac{1}{2}Re\left[iE\left(\frac{1+a}{2},i\right)\right]+\sin(1)+\lim_{y\rightarrow\infty}\int^{y}_{1}\{\sqrt{t}\}\frac{d}{dt}\left(\frac{\sin(t)}{t^{a/2}}\right)dt$$ But $$\int^{y}_{1}\{\sqrt{t}\}\frac{d}{dt}\left(\frac{\sin(t)}{t^{a/2}}\right)dt=\int^{y}_{1}\{\sqrt{t}\}\frac{\cos(t)t^{a/2}-a/2\sin(t)t^{a/2-1}}{t^a}dt=$$ $$\int^{y}_{1}\{\sqrt{t}\}\left(\cos(t)t^{-a/2}-a/2\sin(t)t^{-a/2-1}\right)dt=\int^{y}_{1}\{\sqrt{t}\}\frac{\cos(t)}{t^{a/2}}dt-\frac{a}{2}\int^{y}_{1}\frac{\sin(t)}{t^{a/2+1}}\{\sqrt{t}\}dt.$$ Clearly when $$a$$ is positive and constant $$\lim_{y\rightarrow+\infty}\int^{y}_{1}\frac{\sin(t)}{t^{a/2+1}}\{\sqrt{t}\}dt=2\lim_{x\rightarrow\infty}\int^{x}_{1}\frac{\sin(t^2)}{t^{a+1}}\{t\}dt<\infty,$$ since $$0\leq\{t\}<1$$ and $$-1\leq\sin(t^2)\leq 1$$, for all $$t>0$$.

Hence it remains to find under what condition on $$a>0$$ we have $$\int^{\infty}_{1}\{\sqrt{t}\}\frac{\cos(t)}{t^{a/2}}dt=2\int^{\infty}_{1}\cos(t^2)t^{1-a}\{t\}dt<\infty,$$ knowinig already that for all $$0 we have $$\int^{\infty}_{1}\cos(t^2)t^{1-a}dt<\infty.$$

However it is $$F(x)=\int^{x}_{1}\cos(t)\{\sqrt{t}\}dt=\frac{\sqrt{2\pi}}{2}\textrm{Fs}\left(\sqrt{\frac{2}{\pi}}\right)-\frac{\sqrt{2\pi}}{2}\textrm{Fs}\left(\sqrt{\frac{2x}{\pi}}\right)+$$ $$+\sum_{2\leq k<\sqrt{x}}\sin(k^2)+\{\sqrt{x}\}\sin(x)=O\left(\sum_{2\leq k<\sqrt{x}}\sin(k^2)\right).\tag 2$$ The function $$\textrm{Fs}(x)$$ is the FresnelS function $$\textrm{Fs}(z):=\int^{z}_{0}\sin\left(\pi t^2/2\right)dt$$ and it is known that $$\lim_{x\rightarrow+\infty}\textrm{Fs}(x)=\frac{1}{2}.$$ Hence if we set $$S(x):=\sum_{2\leq k and assume that $$a=1/2-\epsilon$$, $$\epsilon>0$$, then $$I(x)=\int^{x}_{1}\frac{1}{t^{a/2}}\cos(t)\{\sqrt{t}\}dt=\int^{x}_{1}\frac{F'(t)}{t^{a/2}}dt=\frac{F(x)}{x^{a/2}}+\frac{a}{2}\int^{x}_{1}\frac{F(t)}{t^{a/2+1}}dt=S_1+S_2,\tag 3$$ where $$S_1=\frac{1}{x^{a/2}}S(\sqrt{x})+\frac{\{\sqrt{x}\}\sin x}{x^{a/2}}+\frac{\sqrt{\pi/2}}{x^{a/2}}\left(\textrm{Fs}\left(\sqrt{\frac{2}{\pi}}\right)-\textrm{Fs}\left(\sqrt{\frac{2x}{\pi}}\right)\right)$$ and $$S_2=\frac{a}{2}\int^{x}_{1}\frac{1}{t^{1/4-\epsilon/2+1}}\{\frac{\sqrt{2\pi}}{2}\textrm{Fs}\left(\sqrt{\frac{2}{\pi}}\right)-\frac{\sqrt{2\pi}}{2}\textrm{Fs}\left(\sqrt{\frac{2t}{\pi}}\right)+$$ $$+\{\sqrt{t}\}\sin(t) +\sum_{2\leq k\leq \sqrt{t}}\sin(k^2)\}dt.$$ But it is known that there exists constant $$C$$ such that for infinite values of $$x\in\textbf{N}$$ holds $$\sum_{2\leq k\leq x}\sin(k^2)>Cx^{1/2}.\tag 4$$ Hence for infinite values of $$x$$ we will have (easily) $$S_1>C_1x^{\epsilon/2}.\tag 5$$ Moreover if we assume that $$\left|\sum_{2\leq k\leq x}\sin(k^2)\right|=O\left(x^{c+\delta}\right)\textrm{, }\forall \delta>0\textrm{ and }x\rightarrow+\infty,\tag 6$$ then in view of (4) it must be $$c\geq 1/2$$. Also $$S_2=C_{\epsilon}(x)+\int^{x}_{1}t^{-1/4+\epsilon/2-1}\left(\sum_{2\leq k\leq \sqrt{t}}\sin(k^2)\right)dt.$$ Hence $$\left|S_2\right|=\left|C_{\epsilon}(x)+\int^{x}_{1}t^{-1/4+\epsilon/2-1}\left(\sum_{2\leq k\leq \sqrt{t}}\sin(k^2)\right)dt\right|\leq$$ $$\leq \left|\left|C_{\epsilon}(x)\right|+\left|\int^{x}_{1}t^{-1/4+\epsilon/2-1}\left(\sum_{2\leq k\leq \sqrt{t}}\sin(k^2)\right)dt\right|\right|\leq$$ $$|C_{\epsilon}(x)|+\int^{x}_{1}\left|t^{-1/4+\epsilon/2-1}\left(\sum_{2\leq k\leq \sqrt{t}}\sin(k^2)\right)\right|dt=$$ $$=\left|C_{\epsilon}(x)\right|+C_2\int^{x}_{1}t^{-1/4+\epsilon/2-1}\left|\sum_{2\leq k\leq \sqrt{t}}\sin(k^2)\right|dt\leq$$ $$\leq\left|C_{\epsilon}(x)\right|+C_2\int^{x}_1t^{-1-1/4+\epsilon/2}t^{1/4+\delta/2}dt=$$ $$=\left|C_{\epsilon}(x)\right|+C_2\int^{x}_{1}t^{-1+\epsilon/2+\delta/2}dt=\left|C_{\epsilon}(x)\right|+\frac{2}{\delta+\epsilon}\left(x^{(\delta+\epsilon)/2}-1\right)<$$ $$<|C_{0}|+\log x+C_3d\log^2 x,\tag 7$$ where $$\epsilon>0$$ and $$\delta>0$$ so small as we please and $$d=\frac{\epsilon+\delta}{2}>0$$, $$C_3>0$$ constant. It is also easy to see someone that $$\left|C_{\epsilon}(x)\right|$$ are bounded by a constant $$C_0>0$$.

Hence from $$(3)$$ and $$(5),(7)$$ we have if $$a=1/2-\epsilon$$, that $$\left|\int^{x}_{1}\frac{\cos (t)\{\sqrt{t}\}}{t^{a/2}}dt\right|=|S_1+S_2|\geq |S_1|-|S_2|>C_1x^{\epsilon/2}-|C_0|-\log x-C_3d\log^2 x,$$ For infinite values of $$x\in\textbf{N}$$.

Hence $$\lim_{x\rightarrow+\infty}\int^{x}_{1}\frac{\cos(t)\{\sqrt{t}\}}{t^{a/2}}dt=+\infty$$ and we conclude that if (6) holds, then $$\textrm{inf}\geq1/2$$.

I will argue now about the the case $$a=\frac{1}{2}+2\epsilon$$, $$\epsilon>0$$ i.e the case when $$a$$ is not $$1/2$$ but rather a limiting case and doesnot cover the case $$a=1/2$$. Both results $$\textrm{inf}\geq1/2$$ and $$\textrm{inf}=1/2+2\epsilon$$, clearly show us that for $$1/2 the sum $$\sum^{\infty}_{n=1}\frac{\sin(n^2)}{n^a}$$ converges and diverges for $$0, under the hypothesis $$(6)$$. For $$a=1/2$$, we dont know.

For $$a=1/2+2\epsilon$$, $$\epsilon>0$$ and for $$x>>1$$, we chose $$\delta>0$$ such that $$S\left(\sqrt{x}\right)\leq C_1x^{1/4+\delta},$$ we get $$\left|I(x)\right|=|S_1+S_2|\leq \left|C_1\frac{S\left(\sqrt{x}\right)}{x^{a/2}}+C_2\frac{a}{2}\int^{x}_{1}\frac{S\left(\sqrt{t}\right)}{t^{a/2+1}}dt\right|\leq$$ $$\leq\left|C_1'\frac{x^{1/4+\delta}}{x^{1/4+\epsilon}}+C_2'\left(\frac{1}{4}+\epsilon\right)\int^{x}_{1}\frac{t^{1/4+\delta}}{t^{1+1/4+\epsilon}}dt\right|=$$ $$=\left|C_1'x^{-(\epsilon-\delta)}+C_2'\left(\frac{1}{4}+\epsilon\right)\int^{x}_{1}\frac{dt}{t^{1+\epsilon-\delta}}\right|.$$ For $$\delta=\epsilon/2$$ we get $$|I(x)|\leq \left|C_1'x^{-\epsilon/2}-\frac{2C_2'}{\epsilon}\left(\frac{1}{4}+\epsilon\right)\left(x^{-\epsilon/2}-1\right)\right|=$$ $$=\left|C_1'x^{-\epsilon/2}+\frac{C_2'}{2\epsilon}\left(1-x^{-\epsilon/2}\right)+2C_2'\left(1-x^{-\epsilon/2}\right)+H-H\right|\leq$$ $$\leq\left|C_1'x^{-\epsilon/2}+\frac{C_2'}{2\epsilon}\left(1-x^{-\epsilon/2}\right)+2C_2'\left(1-x^{-\epsilon/2}\right)+H\right|+\left|H\right|.\tag 8$$ Now we set $$X=C_1'x^{-\epsilon/2}+\frac{C_2'}{2\epsilon}\left(1-x^{-\epsilon/2}\right)>0$$ and $$Y=2C_2'\left(1-x^{-\epsilon/2}\right)+H>0$$ and I use the inequality $$\left|X+Y\right|\leq \left|X-\frac{Y}{4\epsilon}\right|,\tag 9$$ which is is true for small $$\epsilon$$ and $$x>1$$ since we can write equivalent $$\left|X+Y\right|^2\leq \left|X-\frac{Y}{4\epsilon }\right|^2\Leftrightarrow X^2+Y^2+2XY\leq X^2+\frac{Y^2}{16\epsilon^2}-\frac{XY}{2\epsilon }\Leftrightarrow$$ $$Y^2+2XY\leq\frac{Y^2}{16\epsilon^2}-\frac{XY}{2\epsilon}\Leftrightarrow Y+2X\leq \frac{Y}{16\epsilon^2}-\frac{X}{2\epsilon}\Leftrightarrow$$ $$\left(\frac{1}{16\epsilon^2}-1\right)Y\geq X\left(2+\frac{1}{2\epsilon}\right)$$ This last inequality holds for all small $$\epsilon>0$$ and $$x>>1$$ since it can be writen equivalently as $$(1-16\epsilon^2)Y-X(32\epsilon^2+8\epsilon)\geq0\Leftrightarrow$$ $$2\epsilon(1+2\epsilon)\left(C_2'-4\epsilon C_1'+4C_2'\epsilon\right)x^{-\epsilon/2}\geq 0,$$ where we have used the value $$H=\frac{2(C_2'+4C_2\epsilon)}{1-4\epsilon}$$ Hence (9) is true and we can extract from relation (8) the conclusion $$\left|I(x)\right|\leq \left|C_1'x^{-\epsilon/2}+\frac{C_2'}{2\epsilon}\left(1-x^{-\epsilon/2}\right)+2C_2'\left(1-x^{-\epsilon/2}\right)+H\right|+\left|H\right|=$$ $$=\left|X+Y\right|+\left|H\right|\leq$$ $$\leq\left|C_1'x^{-\epsilon/2}+\frac{C_2'}{2\epsilon}\left(1-x^{-\epsilon/2}\right)-\frac{C_2'}{2\epsilon}\left(1-x^{-\epsilon/2}\right)-\frac{H}{4\epsilon}\right|+|H|=$$ $$=\left|C_1'x^{-\epsilon/2}-\frac{H}{4\epsilon}\right|+|H|.$$ Hence $$\epsilon |I(x)|\leq C_1'x^{-\epsilon/2}\epsilon+H/4+|H|\epsilon.$$ Hence we conclude that $$\epsilon \left|I(x)\right|=O(1)$$ is bounded. Hence for $$\epsilon>0$$ small but constant the $$I(x)$$ are bounded.

• When you applied the partial summation, it seems that you used $C(t)=\sqrt t$. However, $C(t)=\lfloor \sqrt t \rfloor$. – Sungjin Kim Jun 18 '18 at 22:00
• I saw your edit. Then you have to show that the limit as $y\rightarrow\infty$ exists. Otherwise, the proof is incomplete. – Sungjin Kim Nov 13 '18 at 22:46
• You cannot just ignore $\{t\}$ at the last step, since $\cos(t^2)$ changes sign infinitely often. – Sungjin Kim Nov 17 '18 at 8:30
• Yes I know that. Im trying to reduce the problem. – Nikos Bagis Nov 21 '18 at 8:19