Home > Analysis, Fourier Analysis, Fourier Series > Necessary condition for the uniform convergence of a certain series

## Necessary condition for the uniform convergence of a certain series

Let ${a_n}$ be a decreasing sequence of positive numbers such that the series

$\displaystyle \sum_{n = 1}^\infty a_n \sin(nx)$

is uniformly convergent. Then ${(a_n)}$ must satisfy ${(na_n) \rightarrow 0}$.

Note that this result implies that the series ${\sum_{n=1}^\infty a_n \sin(nx)}$ is not uniformly convergent on ${\Bbb{R}}$. It is surprisngly similar to the following result:

Suppose that ${(a_n)}$ is a decreasing sequence of positive real numbers such that the series ${\sum_{n=1}^\infty a_n}$ is convergence. Then ${(na_n) \rightarrow 0}$. It is no surprise that the proofs of these two results are similar.

Proof: If the series converges uniformly, then the following series

$\displaystyle s_k(x)= \sum_{n = [k/2]}^{k} a_n \sin(nx)$

converges uniformly to zero. Picking ${x_k = \frac{\pi}{2k}}$ we find that ${nx_k \in [\pi/4,\pi/2]}$ for ${n = [k/2],...,k}$. Therefore

$\displaystyle a_n \sin(nx_k) \geq \frac{a_n}{\sqrt{2}},n = [k/2],...,k,$

and summing up we get

$\displaystyle s_k(x_k) \geq \frac{a_{[k/2]}+...+a_k}{\sqrt{2}}\geq [k/2]\frac{a_k}{\sqrt{2}}.$

The fact that ${s_k}$ converges uniformly to zero implies that ${(ka_k) \rightarrow 0}$.