Two questions about weakly convergent series related to sin(n2)\sin(n^2) and Weyl’s inequality

By using partial summation and Weyl’s inequality, it is not hard to show that the series n1sin(n2)n is convergent.

  • Is is true that 12=inf
  • 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\}?

Answer

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<a\leq 1 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<x}\sin(k^2)

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<a\leq1 the sum \sum^{\infty}_{n=1}\frac{\sin(n^2)}{n^a} converges and diverges for 0<a<1/2, 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.

Attribution
Source : Link , Question Author : Jack D'Aurizio , Answer Author : Nikos Bagis

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