Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Using Dirichlet series test I've proves that series $\displaystyle\sum\limits_{n=2}^\infty\frac{\sin n x}{n\log n}$ converges for all $x\in\mathbb{R}$.

How to determine whether the series $\displaystyle\sum\limits_{n=2}^\infty\frac{\sin n x}{n\log n}$ converges uniformly or not on $\mathbb{R}$?

share|improve this question
4  
In fact, we can show more generally that if $\{a_n\}$ is a decreasing sequence of real numbers such that $na_n\to 0$ then the series $\sum_n a_n\sin (nx)$ converges uniformly on $\mathbb R$. –  Davide Giraudo Jan 17 '12 at 20:16
    
@DavideGiraudo Will do! –  Pedro Tamaroff Apr 4 '13 at 18:54

1 Answer 1

up vote 6 down vote accepted

Let $\left\{a_n\right\}$ be a decreasing sequence of real numbers such that $n\cdot a_n\to 0$. Then the series $\sum_{n\geqslant 2}a_n\sin (nx)$ 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 $M_n:=\sup_{k\geqslant n}ka_k$, and $R_n(x)=\sum_{k=n}^{\infty}a_k\sin (kx)$. Fix $x\neq 0$ and $N$ such that $\frac 1N\leqslant x<\frac 1{N-1}$. Put for $N>n$: $$A_n(x)=\sum_{k=n}^{N-1}a_k\sin kx\mbox{ and }B_n(x):=\sum_{k=N}^{+\infty}a_k\sin (kx),$$ and for for $n\leqslant N$, $A_n(x)=0$ and the same definition for $B_n$.

Since $|\sin t|\leqslant t$ for $t\geq 0$ we have $$|A_n(x)|\leqslant \sum_{k=n}^{N-1}a_kkx\leqslant M_nx(N-n)\leqslant \frac{N-n}{N-1}M_n,$$ so $|A_n(x)|\leqslant M_n$.

If $N> n$, we have after writing $D_k=\sum_{j=0}^k\sin jx$, $|D_k(x)|\leqslant \frac cx$ on $(0,\delta]$ for some constant $c$. Indeed, we have $|D_k(x)|\leqslant \frac 1{\sqrt{2(1-\cos x)}}$ and $\cos x=1-\frac{x^2}2(1+\xi)$ where $|\xi|\leqslant \frac 12$ so $2(1-\cos x)\geqslant \frac{x^2}2$ and $|D_k(x)\leqslant \frac{\sqrt 2}x$. Therefore $$|B_n(x)|\leqslant \frac{\sqrt 2}x\sum_{k=N}^{+\infty}(a_k-a_{k+1})+a_N\frac{\sqrt 2}x=\frac{2\sqrt 2}xa_N\leqslant 2\sqrt 2 Na_n\leqslant 2\sqrt 2M_n.$$ We get the same bound if $N\leqslant n$. Finally $|R_n(x)|\leqslant (2\sqrt 2+1)M_n$ for all $0\leqslant x\leqslant \delta$, so the convergence is uniform on $\mathbb R$.


Added latter: it's an example of a Fourier series which is uniformly convergent on the real line, but not absolutely convergent at any point of $(0,2\pi)$. Indeed take $x\in (0,2\pi)$. Since $|\sin(nx)|\geqslant \sin^2(nx)$, we would have the convergence of $\sum_{n\geqslant 2}\frac{\sin^2(nx)}{n\log n}$. We have $\sin ^2(nx)= \frac 1{-4}(e^{inx}-e^{-inx})^2=-\frac 14 (e^{2inx}+e^{-2inx}-2)=\frac 12-\frac 12\cos (2nx)$ and an Abel transform shows that the series $\sum_{n\geqslant 2}\frac{\cos(2nx)}{n\log n}$ is convergent. So the series $\sum_{n\geqslant 2}\frac 1{n\log n}$ would be convergent, which is not the case as the integral test shows.

share|improve this answer
    
There some typos. –  no identity Jan 17 '12 at 20:45
    
You're right. I will try to correct the most obvious. You can correct remainding ones –  Davide Giraudo Jan 17 '12 at 20:53
    
Very, nice. Wish I had such a talent! –  no identity Jan 17 '12 at 21:03
    
It's not really talent, since I've already seen this prove in an exercise in book (so there were the step). This series is interesting since we have an example of Fourier series which is uniformly convergent, but it's not absolutely convergent at any point $x\in ]0,2\pi[$. –  Davide Giraudo Jan 17 '12 at 21:07
1  
Another book where you can find the same argument about uniform convergence is N.Bary "A treatise on trigonometric series", Vol.1 pp.87-88, 1964 Pergamon Press. –  Unoqualunque Apr 8 '12 at 7:49

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.