Infinite sum of square root

Question

I tried demonstrate that $\displaystyle \sum_{k=0}^\infty\sqrt{a_n^2+b_n^2}$ is convergent with $a_n$ and $b_n$ are the fourier coefficients of continuos $f(x)$ with $f(x)\in L_2(-\pi, \pi)$ and $f(\pi)=f(-\pi)$. My attempt to show it is to proof that the sum $$\sum_{k=0}^\infty a_n^2+b_n^2$$ is convergent (using Bessel inequation). But I'm not sure if $$\sum_{k=0}^\infty\sqrt{a_n^2+b_n^2}\leq \sum_{k=0}^\infty a_n^2+b_n^2$$

or is it not necessary?

Answer 1

NOT TRUE.

The series $$ f(x)=\sum_{n=1}^\infty \frac{\sin nx}{n} $$ converges conditionally for every $x\in\mathbb R$ and defines a $2\pi-$periodic function $f$, which belongs to $L^2[0,2\pi]$, since $$ \sum_{n=1}^\infty (a_n^2+b_n^2)=\sum_{n=1}^\infty \frac{1}{n^2}<\infty. $$
But $$ \sum_{n=1}^\infty \sqrt{a_n^2+b_n^2}=\sum_{n=1}^\infty \frac{1}{n}=\infty. $$

Answer 2

The inequality holds in the opposite direction with minor changes. Since $\lim_{n\to \infty}a_n=\lim_{n\to \infty}b_n=0$ for $f(x)\in L_2(-\pi,\pi)$, we have some $M$ such that $$n\ge M\implies \sqrt{a_n^2+b_n^2}<1$$hence $$ \sum_{n=M}^\infty {a_n^2+b_n^2}<\sum_{n=M}^\infty \sqrt{a_n^2+b_n^2} $$

Answer 3

An example of a Fourier series whose sum is continuous but the series is not absolutely convergent is

$$ \sum_{n=2}^\infty \frac{\sin(nx)}{n \log n} $$

See e.g. Edwards, "Fourier Series: A Modern Introduction", 7.2.2.