3
$\begingroup$

On each bounded interval $[a,b]$ : $\left|\frac1n \sin\left(\frac xn\right)\right|\le \frac{\max\{|a|,|b|\}}{n^2}$, the series $\sum_{n\ge 1}\frac {\max\{|a|,|b|\}}{n^2}$ converges,

therefore $\sum_{n\ge1}\frac1n \sin\left(\frac xn\right)$ converges uniformly by the Weierstrass M-Test.

May I conclude from this statement that this series converges uniformly on $\mathbb{R}$?

Thanks.

$\endgroup$
  • $\begingroup$ No, from that you cannot conclude uniform convergence on $\mathbb{R}$. The Taylor series of $\sin$ also converges uniformly on every $[a,b]$, but not uniformly on $\mathbb{R}$. $\endgroup$ – Daniel Fischer May 7 '14 at 13:06
  • 2
    $\begingroup$ See the answer here. It could help you. $\endgroup$ – Jika May 7 '14 at 13:06
6
$\begingroup$

The series does not converge uniformly on $\mathbb{R}$.

Consider $x_m = m\cdot \frac{\pi}{2}$ for $m \in \mathbb{Z}^+$. Then

$$\sin \frac{x_m}{n} \geqslant \frac{1}{2}$$

for $m \leqslant n \leqslant 3m$, and so

$$\sum_{n=m}^{3m} \frac{1}{n}\sin \frac{x_m}{n} \geqslant \frac{1}{2} \sum_{n=m}^{3m}\frac{1}{n} > \frac{1}{2} \log 3.$$

$\endgroup$
0
$\begingroup$

Let $\:t_n=1/n,\:\:u_n=\sin\left(\large\frac{x}{n}\normalsize\right)\:$ and $\:S_n(x)=\sum_{j=1}^nt_ju_j,\quad \underbrace{(j,n)\in\mathbb N^2,\:\:j\le n.}_{\text{Property}\{\star\}} $

Therefore, $$S_n(x)=\sum_{j=1}^nt_j(U_j-U_{j-1}),\quad\:U_j=\sum_{k=1}^ju_k\:\{\star\}.\\\implies S_n(x)=t_nU_n+\sum_{j=1}^{n-1}t_{j}U_j-\sum_{j=1}^{n-1}t_{j+1}U_{j-1}=t_nU_n+\sum_{j=1}^{n-1}(t_j-t_{j+1})U_j.$$

Now, $$U_j=\sum_{k=1}^n\sin(x/k)=\mathcal{Im}\left(\sum_{k=1}^ne^{ix/k}\right)=\mathcal{Im}\left(\frac{e^{ix}-e^{ix/(n+1)}}{1-e^{ix}}\right)=\mathcal{Im}\left(\frac{e^{ix}-e^{ix/(n+1)}}{-2i\sin\left(\frac{x}{2}\right)e^{ix/2}}\right).$$

$\implies\forall x\in\{0,2\pi\mathbb Z\},\:\:S_n(x)\:$ becomes unbounded as $\:n\to\infty$.

$\endgroup$
0
$\begingroup$

Choose a $\:x\:$ that will create a contradiction such as $\:x=n.$

$$\implies \sum_{n\in\mathbb N}\frac{1}{n}\sin(1)\overset{\text{Harmonic}}{=}\infty$$ So that it does not converge uniformly on any bounded set since it doesn not converge on $\:\mathbb R_+.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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