Jtdylktuy

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

• Is is true that $$frac{1}{2}=infleft{alphainmathbb{R}^+:sum_{ngeq 1}frac{sin(n^2)}{n^alpha}mbox{ is convergent}right}?$$


• In the case of a positive answer to the previous question, what is $$infleft{betainmathbb{R}^+:sum_{ngeq 1}frac{sin(n^2)}{sqrt{n}(log n)^beta}mbox{ is convergent}right}?$$

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 $xgeq lambda_1$. Put
$$
C(x)=sum_{lambda_nleq x}c_n,
$$
where the summation is over all $n$ for which $lambda_nleq x$. Then for $xgeqlambda_1$,
$$
sum_{lambda_nleq x}c_nf(lambda_n)=C(x)f(x)-int^{x}_{lambda_1}C(t)f'(t)dt.tag 1
$$

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

Hence when $a>0$, then
$$
lim_{xrightarrowinfty}sum^{x}_{n=1}frac{sin(n^2)}{n^a}=frac{1}{2}Releft[iEleft(frac{1+a}{2},iright)right]+sin(1)+lim_{yrightarrowinfty}int^{y}_{1}{sqrt{t}}frac{d}{dt}left(frac{sin(t)}{t^{a/2}}right)dt
$$
But
$$
int^{y}_{1}{sqrt{t}}frac{d}{dt}left(frac{sin(t)}{t^{a/2}}right)dt=int^{y}_{1}{sqrt{t}}frac{cos(t)t^{a/2}-a/2sin(t)t^{a/2-1}}{t^a}dt=
$$
$$
int^{y}_{1}{sqrt{t}}left(cos(t)t^{-a/2}-a/2sin(t)t^{-a/2-1}right)dt=int^{y}_{1}{sqrt{t}}frac{cos(t)}{t^{a/2}}dt-frac{a}{2}int^{y}_{1}frac{sin(t)}{t^{a/2+1}}{sqrt{t}}dt.
$$
Clearly
$$
int^{y}_{1}frac{sin(t)}{t^{a/2+1}}{sqrt{t}}dt=2int^{x}_{1}frac{sin(t^2)}{t^{a+1}}{t}dt<infty,
$$
for all $a>0$, since $0leq{t}<1$ and $-1leqsin(t^2)leq 1$, for all $t>0$.

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