On the convergence of $\sum a_n \cos nx$ , $\forall x \in \mathbb R \setminus \pi \mathbb Z$ , where $\{a_n\}$ is monotone

Question

Let $\{a_n\}$ be a monotone sequence of real numbers ; if $\lim_{ n \to \infty} a_n=0$ , then is it true that $\sum a_n \cos nx$ is convergent $\forall x \in \mathbb R \setminus \pi \mathbb Z$ ?Is the convergence of the series uniform over $\mathbb R \setminus \pi \mathbb Z$ ? Is the converse true i.e if $\{a_n\}$ be a monotone sequence of real numbers such that $\sum a_n \cos nx$ is convergent $\forall x \in \mathbb R \setminus \pi \mathbb Z$ , then is it true that $a_n \to 0$ ?

Answer 1

Yes - in fact the slightly stronger statement is true: $\sum a_n e^{inx}$ is convergent (the given sum is the real part of this). WLOG the $a_n$ are positive and decreasing. Summation by parts gives, for $x\notin 2\pi\Bbb Z$, \begin{align*} \bigg| \sum_{n=M}^N a_n e^{inx} \bigg| &= \bigg| \sum_{n=M}^{N-1} \bigg( \sum_{k=M}^n e^{ikx} \bigg) (a_n-a_{n+1}) + \bigg( \sum_{k=M}^N e^{ikx} \bigg) a_N \bigg| \\ &= \bigg| \sum_{n=M}^{N-1} \frac{e^{i(n+1)x}-e^{iMx}}{e^{ix}-1} (a_n-a_{n+1}) + \frac{e^{i(N+1)x}-e^{iMx}}{e^{ix}-1} a_N \bigg| \\ &\le \frac2{|e^{ix}-1|} \bigg( \sum_{n=M}^{N-1} (a_n-a_{n+1}) + a_N \bigg) = \frac2{|e^{ix}-1|}a_M. \end{align*} Since $a_M\to0$, this can be made less than $\varepsilon$ by choosing $M$ large enough. Therefore the series $\sum_{n=1}^\infty a_n e^{inx}$ satisfies the Cauchy criterion and hence converges.

The convergence is not necessarily uniform: $a_n=\frac1n$ is a counterexample (the resulting series is unbounded for $x$ near $2\pi\Bbb Z$).

I believe the converse is true by the following argument: if $a_n\not\to0$ then $a_n\to L$ for some $L\neq0$ (since $\{a_n\}$ is monotone); then $\sum a_n\cos nx = \sum(a_n-L)\cos nx + L \sum \cos nx$ is the sum of a series that converges for all $x\notin 2\pi\Bbb Z$ and a series that doesn't converge anywhere (since its terms don't tend to $0$), hence converges for no $x\in 2\pi\Bbb Z$. One has to deal with the case $|a_n|\to\infty$ as well....

Answer 2

A bit too long for a comment: $\sum_1^N \cos(nx)=\dfrac{\sin(\frac{Nx}{2})\cos(\frac{1}{2}(N+1)x)}{\sin(\frac{x}{2})}$ (see here). Thus the partial sums are bounded for $x\not = k\pi$ for some $k\in\mathbb{Z}$. Moreover, since $a_n$ is monotone decreasing, then the result of the first question I think follows from Dirichlet's Test.