Cfrgtkky

Study the convergence of $$displaystyle sum_{ngeq 1} frac{(-1)^n|sin n|}{n}.$$

I am stuck with this series, we need probably some measure of irrationally of $pi$, unfortunately I am unfamiliar with this. So here is my attempt :

Let $f(x) = sum frac{|sin{n}|}{n} x^n, |x| < 1$

It's not difficult to compute the Fourier series of $|sin(x)|$ :

$$
displaystyle|sin(x)|=frac{2}{pi}-frac{4}{pi}sum_{n=1}^{+infty}frac{cos(2nx)}{4n^2-1}
$$

Then Fubini's theorem (Series Version) works very well (because the previous series converges absolutely at $x$ fixed ) and all calculations made, we find that for all $xin( -1,1)$:

$$
displaystyle f(x)=frac{2}{pi}sum_{n=1}^{+infty}frac{x^n}{n}-frac{4}{pi}sum_{p=1}^{+infty}frac{x^2-2xcos(p)}{(4p^2-1)(x^2-2xcos(p)+1)}
$$

However, the second sum I have not been able to show the convergence. I feel the series diverge because the following series
$$
displaystylesumfrac{1}{p^2sin^2left(frac{p}{2}right)}
$$

diverge because $0$ is an accumulation point of $displaystyle (nsin(n))$ sequence.

Any ideas (for the original series) ?

By Dirichlet's convergence test, this series will converge if we can show that there exists a constant $C$ such that $$left|sum_{nleq x}(-1)^{n}|sin(n)|right|leq C$$ for all $x$.

Lets write $$sum_{nleq x}(-1)^{n}|sin(n)|=sum_{nleqfrac{x}{2}}|sin(2n)|-sum_{nleqfrac{x+1}{2}}|sin(2n-1)|.$$ Then for $x=2N$, an even number, Euler Maclaurin summation yields $$sum_{nleq N}|sin(2n)|=int_{1}^{N}|sin(2t)|dt+sum_{k=1}^{K}frac{(-1)^{k}}{k!}B_{k}left(frac{d^{k-1}}{dt^{k-1}}|sin(2t)|biggr|_{t=1}^{t=N}right)$$ $$ -frac{(-1)^{K}}{K!}int_{1}^{N}B_{K}({t})left(frac{d^{k}}{dt^{k}}|sin(2t)|right)dt.$$ Note that $|sin(x)|$ has infinitely many derivatives everywhere except at integer multiples of $pi$, and so the above holds for any $K>0$. Since $$|B_{k}({x})|leq k!2^{1-k}pi^{-k}zeta(k),$$ and since the derivatives of $|sin(t)|$ are bounded in absolute value by $1$, it follows that $$left|sum_{nleq N}|sin(2n)|-int_{1}^{N}|sin(2t)|dtright|leq4sum_{k=1}^{K}frac{zeta(k)}{(2pi)^{k}}+frac{2zeta(K)N}{(2pi)^{K}}.$$ The series $sum_{k=1}^{infty}frac{zeta(k)}{(2pi)^{k}}$ converges absolutely, so by taking $K=N$ we see that there exists a constant $C_{1}$ such that $$left|sum_{nleq N}|sin(2n)|-int_{1}^{N}|sin(2t)|dtright|leq C_{1}$$ for all $N$. Similarly, there exists a constant $C_{2}$ such that $$left|sum_{nleq N}|sin(2n-1)|-int_{1}^{N}|sin(2t-1)|dtright|leq C_{2}.$$ Thus by the triangle inequality, $$left|sum_{nleq x}(-1)^{n}|sin(n)|right|leq C_{1}+C_{2}+left|int_{1}^{N}|sin(2t)|dt-int_{1}^{N}|sin(2t-1)|dtright|$$

$$leq C_{1}+C_{2}+int_{N-1/2}^{N}|sin(2t)|dt+int_{1}^{3/2}|sin(2t-1)|dt$$

$$leq C_{3}$$ for some constant $C_{3}$. This implies the desired result.

The series converges. It is enough to show that the sequence of the following partial sums converges:
begin{align}
s_N &= sum_{n=1}^{N} left(frac{ |sin 2n|}{2n}-frac{|sin (2n+1)|}{2n+1}right)=sum_{n=1}^N left(frac{(2n+1)|sin 2n|- 2n|sin (2n+1)| }{2n(2n+1)} right)\
&=sum_{n=1}^N left( frac{|sin 2n|-|sin (2n+1)|}{2n+1}+frac{|sin 2n|}{2n(2n+1)}right).
end{align}
Thus, it is enough to show that the following converges:
$$
S_N = sum_{n=1}^N frac{|sin 2n|-|sin (2n+1)|}{2n+1}.
$$
We consider a partition of $mathbb{N}$ into four disjoint sets $A_{1}$, $A_{2}$, $A_{3}$, $A_{4}$ defined by:
$$
A_{1}={ninmathbb{N}: sin 2n >0, sin (2n+1)>0}, A_{2}={ninmathbb{N}: sin 2n >0, sin (2n+1)<0},
$$
$$
A_{3}={ninmathbb{N}: sin 2n <0, sin (2n+1)>0}, A_{4}={ninmathbb{N}: sin 2n <0, sin (2n+1)<0}.
$$
Note that
$$
A_{1}={ninmathbb{N}: 2n mathrm{mod} 2pi in (0,pi-1)}, A_{2}={ninmathbb{N}: 2n mathrm{mod} 2pi in (pi-1,pi)},$$
$$
A_{3}={ninmathbb{N}: 2n mathrm{mod} 2pi in (-1,0)}, A_{4}={ninmathbb{N}: 2n mathrm{mod} 2pi in (-pi, -1)}.$$
By trigonometric identities,
$$
nin A_{1} Longrightarrow |sin 2n|-|sin (2n+1)| = sin 2n - sin(2n+1) = -2cos(2n+frac12)sin frac12, $$
$$
nin A_{2} Longrightarrow |sin 2n|-|sin (2n+1)| = sin 2n + sin(2n+1) = 2sin(2n+frac12)cos frac12, $$
$$
nin A_{3} Longrightarrow |sin 2n|-|sin (2n+1)| = -sin 2n - sin(2n+1) = -2sin(2n+frac12)cosfrac12, $$
$$
nin A_{4} Longrightarrow |sin 2n|-|sin (2n+1)| = -sin 2n + sin(2n+1) = 2cos(2n+frac12)sin frac12.$$
We define
$$
f_1(x)=-I_{(0,pi-1)}(x)2cos(x+frac12)sinfrac12, f_2(x)=I_{(pi-1,pi)}(x)2sin(x+frac12)cosfrac12,$$
$$
f_3(x)=-I_{(-1,0)}(x)2sin(x+frac12)cosfrac12, f_4(x)=I_{(-pi,-1)}(x)2cos(x+frac12)sinfrac12 $$
where $I_A$ is the characteristic function of $A$. Note that these functions $f_i(x)$ are of bounded variation in $[-pi,pi]$. Thus, $f=f_1+f_2+f_3+f_4$ is of a bounded variation.

We need Koksma's inequality p. 143, Theorem 5.1 of 'Uniform Distribution of Sequences' by Kuipers and Niederreiter:

Theorem [Koksma]

Let $f$ be a function on $I=[0,1]$ of bounded variation $V(f)$, and suppose we are given $N$ points $x_1, ldots , x_N$ in $I$ with discrepancy
$$
D_N:=sup_{0leq aleq bleq 1} left|frac1N #{1leq nleq N: x_n in (a,b) } -(b-a)right|.
$$
Then
$$
left|frac1N sum_{nleq N} f(x_n) - int_I f(x)dx right|leq V(f)D_N.
$$

To control the discrepancy, we apply Erdos-Turan inequality p. 112, Theorem 2.5 of Kuipers and Niederreiter's book.

Theorem[Erdos-Turan]

Let $x_1, ldots, x_N$ be $N$ points in $I=[0,1]$. Then there is an absolute constant $C>0$ such that for any positive integer $m$,
$$
D_Nleq C left( frac1m+ sum_{h=1}^m frac1h left| frac1Nsum_{n=1}^N e^{2pi i h x_n}right|right).
$$

The sequence of our interest is $x_n = 2n$ mod $2pi$. The above two inequalities together applied to $f=f_1+f_2+f_3+f_4$ yields the following: There is an absolute constant $C>0$ such that for any positive integer $m$,
$$
left|frac1N sum_{nleq N} f(x_n) - frac1{2pi} int_{-pi}^{pi} f(x)dx right|leq C left( frac1m+ frac1Nsum_{h=1}^m frac1{hlangle frac h{pi} rangle}right).
$$
A result on the irrationality measure of $pi$ by Salikhov implies that
$$
left| frac1{pi} - frac pq right| geq frac 1{q^{mu+epsilon}}
$$
for all integers $p, q$ and $q$ is sufficiently large, and $mu=7.60631$, $epsilon>0$. This implies
$$
hleftlangle frac h{pi} rightrangle geq h^{2-mu-epsilon}
$$
for sufficiently large $h$. Then for some absolute constant $C>0$,
$$
left|frac1N sum_{nleq N} f(x_n) - frac1{2pi} int_{-pi}^{pi} f(x)dx right|leq C left( frac1m+ frac1N m^{mu-1+epsilon}right).
$$
Taking $m=lfloor N^{1/mu}rfloor$, we obtain
$$
left|frac1N sum_{nleq N} f(x_n) - frac1{2pi} int_{-pi}^{pi} f(x)dx right|leq C N^{-frac1{mu}+epsilon}.
$$
It is easy to see that $int_{-pi}^{pi} f(x) dx = 0$. Therefore,
$$
left|sum_{nleq N}f(x_n)right|leq CN^{1-frac1{mu}+epsilon}.
$$
The convergence of the series now follows from Abel's summation formula.

Added on 12/1/2018

By the bound for the discrepancy here: https://en.wikipedia.org/wiki/Low-discrepancy_sequence#Additive_recurrence

We may have a better error term:
$$
left|sum_{nleq N}f(x_n)right|leq CN^{1-frac1{mu-1}+epsilon}.
$$

$color{red}{text{Not an answer, just an idea needing more work}}$ :

I'd say $sum_n frac{(-1)^n}{n} b_n$ converges whenever $Delta^k b(n) = O(w^{k}),w< 2$ where $Delta^k b(n)$ is the $k$th forward difference, here $b(n) = |sin n|$

If you sum by parts $k$ times, using that $sum_{n=1}^N frac{(-1)^n}{n} = frac{(-1)^N}{2 N}+ O(frac{1}{2 N^2})$ you'll get a main term $2^{-k} sum_{n=1}^{N-k} frac{(-1)^n}{n} Delta^k b(n)$

use that $A(2M)=sum_{n=M}^infty frac{1}{(2n+1)(2n+2)} = sum_{n=m}^infty frac{1}{(2n+1)^2}-frac{1}{(2n+1)^2(2n+2)}$ and approximate with $int_{2M}^infty frac{dx}{x^2}-int_{2M}^infty frac{dx}{x^3}$ to obtain

$$A(N) = ln 2+sum_{n=1}^N frac{(-1)^n}{n}=frac{(-1)^N}{2N}+O(frac{1}{2 N^2})$$

Summing by parts
$$sum_{n=1}^N frac{(-1)^n}{n} |sin n| = A(N)|sin(N)|+sum_{n=1}^{N-1} A(n) (|sin n| - |sin (n +1)|)$$

The problematic term is $sum_{n=1}^{N-1} frac{(-1)^N}{2N} (|sin n| - |sin (n +1)|) $
that we can sum by parts again to get a new problematic term
$$sum_{n=1}^{N-2} frac{(-1)^n}{4n} Delta^2 b(n)$$
where $Delta^2 b(n)=(|sin n| - |sin( n +1)|)-(|sin( n+1)| - |sin( n +2)|)$

summing by parts $k$ times we'll have
$$frac{1}{2^k}sum_{n=1}^{N-k} frac{(-1)^n}{n} Delta^k b(n)$$
Where $Delta^k b(n)$ is the $k$th forward difference of $b(n) = |sin n|$

By using the Fourier series of $left|sin(n)right|$ the problem of showing that
$$ sum_{ngeq 1}frac{(-1)^n left|sin nright|}{n} $$
is convergent boils down to the problem of showing that
$$ sum_{mgeq 1}frac{logleft|cos mright|}{4m^2-1}$$
is convergent. The only issue is given by the values of $m$ such that $m$ is close to an odd multiple of $frac{pi}{2}$.

On the other hand, since the irrationality measure of $pi$ is finite, even if $cos m$ is close to zero it cannot be closer than $frac{1}{m^{10}}$ for any $m$ large enough. Since
$$ sum_{mgeq 1}frac{10log m}{4m^2-1} $$
is absolutely convergent, the original series is convergent as well.