Infinite sum of square root 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?

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
$$
\sum_{n=1}^\infty \sqrt{a_n^2+b_n^2}=\sum_{n=1}^\infty \frac{1}{n}=\infty.
$$
