3
$\begingroup$

Let $\{a_n\}$ be decreasing sequence of positive terms then prove that $\displaystyle \sum a_n \sin (nx)$ converges uniformly on $\Bbb{R}$ iff $na_n \to 0$ as $n\to \infty$.

I proved that convergence of $ \sum a_n \sin (nx)$ implies $na_n \to 0$ as $n\to \infty$ . I got stuck while prove the converse, I tried ussing Dirichlet's test but the problem here is that partial sums of $\displaystyle \sum^{n}_{k=1} \sin (kx)$ are bounded by $\displaystyle \frac{1}{|\sin (\frac{t}{2})|}$ , so as per Dirichlet's test requirement I'm not getting uniform bound.

$\endgroup$
  • $\begingroup$ Could you please post your solution for the convergence of $\sum a_n \sin(nx) \implies n a_n\to 0$? $\endgroup$ – Ozera Feb 25 '15 at 22:35
1
$\begingroup$

You can try to write :

$a_n*sin(n*x) = n*a_n*\frac{sin(n*x)}{n}$

Prove that the series $\sum \frac{sin(n*x)}{n}$ converges, I think you can even have good informations about its limit, by calculating the partial sum of the cos(n*x) and integrating it

with this and the condition n*$a_n$ -> 0, you can find something

$\endgroup$
0
$\begingroup$

If: $$ D_N(x) = \sum_{n=1}^{N}\sin(nx) $$ then we have: $$\left| D_N(x) \right|\leq \min\left(N,\frac{1}{|\sin(x/2)|}\right) $$ and the claim follows by partial summation.

$\endgroup$

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.