 # Using Dirichlet series test I proved that the series <mstyle displaystyle="true" scriptlevel="0"> Logan Wyatt 2022-07-04 Answered
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 $x\in \mathbb{R}$
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Let $\left\{{a}_{n}\right\}$ be a decreasing sequence of real numbers such that $n\cdot {a}_{n}\to 0$. Then the series $\sum _{n⩾2}{a}_{n}\mathrm{sin}\left(nx\right)$ is uniformly convergent on $\mathbb{R}$
Thanks to Abel transform, we can show that the convergence is uniform on $\left[\delta ,2\pi -\delta \right]$ for all $\delta >0$. Since the functions are odd, we only have to prove the uniform convergence on $\left[0,\delta \right]$. Put ${M}_{n}:=\underset{k⩾n}{sup}k{a}_{k}$, and ${R}_{n}\left(x\right)=\sum _{k=n}^{\mathrm{\infty }}{a}_{k}\mathrm{sin}\left(kx\right)$. Fix $x\ne 0$ and N such that $\frac{1}{N}⩽x<\frac{1}{N-1}$. Put for $n

and for ${A}_{n}\left(x\right):=0$
Since $|\mathrm{sin}t|⩽t$ for $t\ge 0$ we have
$|{A}_{n}\left(x\right)|⩽\sum _{k=n}^{N-1}{a}_{k}kx⩽{M}_{n}x\left(N-n\right)⩽\frac{N-n}{N-1}{M}_{n},$
so $|{A}_{n}\left(x\right)|⩽{M}_{n}$
If $N>n$, we have after writing ${D}_{k}=\sum _{j=0}^{k}\mathrm{sin}jx$ on $\left(0,\delta \right]$ for some constant c. Indeed, we have $|{D}_{k}\left(x\right)|⩽\frac{1}{\sqrt{2\left(1-\mathrm{cos}x\right)}}$ and $\mathrm{cos}x=1-\frac{{x}^{2}}{2}\left(1+\xi \right)$ where $|\xi |⩽\frac{1}{2}$ so $2\left(1-\mathrm{cos}x\right)⩾\frac{{x}^{2}}{2}$ and $|{D}_{k}\left(x\right)⩽\frac{\sqrt{2}}{x}$. Therefore
$|{B}_{n}\left(x\right)|⩽\frac{\sqrt{2}}{x}\sum _{k=N}^{+\mathrm{\infty }}\left({a}_{k}-{a}_{k+1}\right)+{a}_{N}\frac{\sqrt{2}}{x}=\frac{2\sqrt{2}}{x}{a}_{N}⩽2\sqrt{2}N{a}_{N}⩽2\sqrt{2}{M}_{n}.$

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