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.

  • $\begingroup$ 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

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


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.


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