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and

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

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

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

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

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