Mathematical Analysis Fundamentals

and

${\beta }_n=\frac{1}{\pi }{\int }_{-\pi }^{\pi }{f}^{\prime }(x)\sin \mathit{nx}\mathit{dx}=-\frac{n}{\pi }{\int }_{-\pi }^{\pi }f(x)\cos \mathit{nx}\mathit{dx}=-{\mathit{na}}_n\text{.}$

Finally, by Theorem 4.7, Corollary 12.33, and Example 12.15,

$\begin{array}{cc}\hfill {\left({\displaystyle \sum _{n=1}^m}\sqrt{a_n^2+b_n^2}\right)}^2=& {\left({\displaystyle \sum _{n=1}^m}\frac{1}{n}\sqrt{{\alpha }_n^2+{\beta }_n^2}\right)}^2\hfill \\ \hfill \le & \left({\displaystyle \sum _{n=1}^m}\frac{1}{n^2}\right)\left({\displaystyle \sum _{n=1}^m}({\alpha }_n^2+{\beta }_n^2)\right)\le \frac{\pi }{6}{\int }_{-\pi }^{\pi }{f}^{\prime }{(x)}^2\mathit{dx}\text{.}\hfill \end{array}$

Therefore, the sequence of partial sums of the series ${\sum }_{n=1}^{\infty }\sqrt{a_n^2+b_n^2}$ is bounded and increasing, respectively, and it converges.  $\blacksquare$

Theorem 12.46

Let $f\in C(-\pi \text{,}\pi )\cup \mathit{PS\; ...}$