# Fourier series convergence without sines or cosines converging

Is it possible to have a Fourier series $$a_0 + \sum_{k=1}^{\infty} \left[a_k\cos(kx) + b_k\sin(kx)\right]$$ converge without either the cosines or the sines converging?

Here is my work so far: Since the series converges for all $x$, it has to converge when $x = \frac{\pi}{2}$. Then $\sum_{k=1}^{\infty}b_k$ converges.

But I don't know how to go on from here.

• Could you check the value of the series for x = Pi / 2 ? I have the feeling that some a(k) will stay and that some b(k) will disappear. – Claude Leibovici Dec 1 '13 at 6:30

Let $$f(x) = \lim_{n\to\infty} a_0 + \sum_{k=1}^n a_k \cos(kx) + b_k \sin(kx) .$$ Then we also have $$\frac12(f(x)+f(-x)) = \lim_{n\to\infty} a_0 + \sum_{k=1}^n a_k \cos(kx) ,$$ and $$\frac12(f(x)-f(-x)) = \lim_{n\to\infty} a_0 + \sum_{k=1}^n b_k \sin(kx) ,$$ This works no matter which kind of limit you use (e.g. pointwise, a.e., in $L^p$, etc). You can even use different summation methods (e.g. Cesàro).