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}.

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