Logan Wyatt

2022-07-04

Using Dirichlet series test I proved that the series $\sum _{n=2}^{\mathrm{\infty}}\frac{\mathrm{sin}nx}{n\mathrm{log}n}$ converges for all $x\in \mathbb{R}$

engaliar0l

Beginner2022-07-05Added 13 answers

Let $\left\{{a}_{n}\right\}$ be a decreasing sequence of real numbers such that $n\cdot {a}_{n}\to 0$. Then the series $\sum _{n\u2a7e2}{a}_{n}\mathrm{sin}(nx)$ is uniformly convergent on $\mathbb{R}$

Thanks to Abel transform, we can show that the convergence is uniform on $[\delta ,2\pi -\delta ]$ for all $\delta >0$. Since the functions are odd, we only have to prove the uniform convergence on $[0,\delta ]$. Put ${M}_{n}:=\underset{k\u2a7en}{sup}k{a}_{k}$, and ${R}_{n}(x)=\sum _{k=n}^{\mathrm{\infty}}{a}_{k}\mathrm{sin}(kx)$. Fix $x\ne 0$ and N such that $\frac{1}{N}\u2a7dx<\frac{1}{N-1}$. Put for $n<N$

${A}_{n}(x)=\sum _{k=n}^{N-1}{a}_{k}\mathrm{sin}kx{\textstyle \text{and}}{B}_{n}(x):=\sum _{k=n}^{+\mathrm{\infty}}{a}_{k}\mathrm{sin}(kx),$

and for ${A}_{n}(x):=0$

Since $|\mathrm{sin}t|\u2a7dt$ for $t\ge 0$ we have

$|{A}_{n}(x)|\u2a7d\sum _{k=n}^{N-1}{a}_{k}kx\u2a7d{M}_{n}x(N-n)\u2a7d\frac{N-n}{N-1}{M}_{n},$

so $|{A}_{n}(x)|\u2a7d{M}_{n}$

If $N>n$, we have after writing ${D}_{k}=\sum _{j=0}^{k}\mathrm{sin}jx$ on $(0,\delta ]$ for some constant c. Indeed, we have $|{D}_{k}(x)|\u2a7d\frac{1}{\sqrt{2(1-\mathrm{cos}x)}}$ and $\mathrm{cos}x=1-\frac{{x}^{2}}{2}(1+\xi )$ where $|\xi |\u2a7d\frac{1}{2}$ so $2(1-\mathrm{cos}x)\u2a7e\frac{{x}^{2}}{2}$ and $|{D}_{k}(x)\u2a7d\frac{\sqrt{2}}{x}$. Therefore

$|{B}_{n}(x)|\u2a7d\frac{\sqrt{2}}{x}\sum _{k=N}^{+\mathrm{\infty}}({a}_{k}-{a}_{k+1})+{a}_{N}\frac{\sqrt{2}}{x}=\frac{2\sqrt{2}}{x}{a}_{N}\u2a7d2\sqrt{2}N{a}_{N}\u2a7d2\sqrt{2}{M}_{n}.$

Thanks to Abel transform, we can show that the convergence is uniform on $[\delta ,2\pi -\delta ]$ for all $\delta >0$. Since the functions are odd, we only have to prove the uniform convergence on $[0,\delta ]$. Put ${M}_{n}:=\underset{k\u2a7en}{sup}k{a}_{k}$, and ${R}_{n}(x)=\sum _{k=n}^{\mathrm{\infty}}{a}_{k}\mathrm{sin}(kx)$. Fix $x\ne 0$ and N such that $\frac{1}{N}\u2a7dx<\frac{1}{N-1}$. Put for $n<N$

${A}_{n}(x)=\sum _{k=n}^{N-1}{a}_{k}\mathrm{sin}kx{\textstyle \text{and}}{B}_{n}(x):=\sum _{k=n}^{+\mathrm{\infty}}{a}_{k}\mathrm{sin}(kx),$

and for ${A}_{n}(x):=0$

Since $|\mathrm{sin}t|\u2a7dt$ for $t\ge 0$ we have

$|{A}_{n}(x)|\u2a7d\sum _{k=n}^{N-1}{a}_{k}kx\u2a7d{M}_{n}x(N-n)\u2a7d\frac{N-n}{N-1}{M}_{n},$

so $|{A}_{n}(x)|\u2a7d{M}_{n}$

If $N>n$, we have after writing ${D}_{k}=\sum _{j=0}^{k}\mathrm{sin}jx$ on $(0,\delta ]$ for some constant c. Indeed, we have $|{D}_{k}(x)|\u2a7d\frac{1}{\sqrt{2(1-\mathrm{cos}x)}}$ and $\mathrm{cos}x=1-\frac{{x}^{2}}{2}(1+\xi )$ where $|\xi |\u2a7d\frac{1}{2}$ so $2(1-\mathrm{cos}x)\u2a7e\frac{{x}^{2}}{2}$ and $|{D}_{k}(x)\u2a7d\frac{\sqrt{2}}{x}$. Therefore

$|{B}_{n}(x)|\u2a7d\frac{\sqrt{2}}{x}\sum _{k=N}^{+\mathrm{\infty}}({a}_{k}-{a}_{k+1})+{a}_{N}\frac{\sqrt{2}}{x}=\frac{2\sqrt{2}}{x}{a}_{N}\u2a7d2\sqrt{2}N{a}_{N}\u2a7d2\sqrt{2}{M}_{n}.$

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