[jacks]

\( \displaystyle \frac{\cos(2a)}{1\cdot 3}+\frac{\cos(4a)}{3\cdot 5}+\frac{\cos(6a)}{5\cdot 7}+\cdots \cdots  \)

[Richard R Richard]

\( f(a)=\displaystyle \sum \limits_{x=1}^{\infty}\dfrac{\cos(2xa)}{(2x-1)(2x+1)}= \sum \limits_{x=1}^{\infty}\dfrac{\cos(2xa)}{4x^2-1}=... \)

Its result depends of the value of $$a$$.
If $$a=k\pi/2+\pi/4$$ with $$k\in \mathbb N$$ then the sum $$f= 0$$.

and

$$f(0)=-f(\pi/2)=1/2$$

Allways

$$\displaystyle \sum \limits_{x=1}^{\infty}\dfrac{1}{4x^2-1}$$ converge to $$1/2$$

[Masacroso]

\( \displaystyle \frac{\cos(2a)}{1\cdot 3}+\frac{\cos(4a)}{3\cdot 5}+\frac{\cos(6a)}{5\cdot 7}+\cdots \cdots  \)

If there is no mistake somewhere

\( \displaystyle{
\begin{align*}
\sum_{k\geqslant 1}\frac{\cos (2ka)}{(2k+1)(2k-1)}&=\operatorname{Re}\sum_{k\geqslant 1}\frac{z^{2k}}{(2k+1)(2k-1)},\quad z:=e^{ia}\\
&=\operatorname{Re}\lim_{r\to 1^-}\sum_{k\geqslant 1}\frac{\omega ^{2k}}{(2k+1)(2k-1)},\quad \omega :=rz\\
&=\frac1{2}\operatorname{Re}\lim_{r\to 1^-}\sum_{k\geqslant 1}\omega ^{2k}\left(\frac1{2k-1}-\frac1{2k+1}\right)\\
&=\frac1{2}\operatorname{Re}\lim_{r\to 1^-}\left( \sum_{k\geqslant 0}\frac{\omega ^{2k+2}}{2k+1}-\sum_{k\geqslant 1}\frac{\omega ^{2k}}{2k+1}\right)\\
&=\frac1{2}\operatorname{Re}\lim_{r\to 1^-}\left( \omega ^2+(\omega ^2-1)\sum_{k\geqslant 1}\frac{\omega ^{2k}}{2k+1}\right)\\
&=\frac1{2}\operatorname{Re}\lim_{r\to 1^-}\left( \omega ^2+\frac{\omega ^2-1}{\omega }\sum_{k\geqslant 1}\frac{\omega ^{2k+1}}{2k+1}\right)\\
&=\frac1{2}\operatorname{Re}\lim_{r\to 1^-}\left( \omega ^2+\frac{\omega ^2-1}{\omega }\left(\sum_{k\geqslant 2}\frac{\omega ^k}{k}-\sum_{k\geqslant 1}\frac{\omega ^{2k}}{2k}\right)\right)\\
&=\frac1{2}\operatorname{Re}\lim_{r\to 1^-}\left( \omega ^2+\frac{\omega ^2-1}{\omega }\left(-\omega -\ln (1-\omega )+\frac1{2}\ln(1-\omega ^2)\right)\right)\\
&=\frac1{2}\operatorname{Re}\lim_{r\to 1^-}\left( 1+\frac{\omega ^2-1}{\omega }\cdot \frac1{2}\ln \left(\frac{1+\omega }{1-\omega }\right)\right)\\
&=\frac1{2}+\operatorname{Re}\left(\frac1{4}(z-\bar z)\cdot \lim_{r\to 1^-}\ln \left(\frac{1+\omega }{1-\omega }\right)\right)\\
&=\frac1{2}-\frac1{2}\operatorname{sen}(a)\lim_{r\to 1^-}\arg\left(\frac{1+ re^{ia}}{1-re^{ia}}\right)\\
&=\frac1{2}-\frac{\pi}{4}|\operatorname{sen}(a)|
\end{align*}
} \)

because

\( \displaystyle{
\lim_{r\to 1^-}\arg\left(\frac{1+ re^{ia}}{1-re^{ia}}\right)=\begin{cases}
\pi/2,& a\in(0,\pi)\\
-\pi/2,& a\in(-\pi,0)\\
0,&\text{ otherwise }
\end{cases}
} \)

[Abdulai]

\( \displaystyle \frac{\cos(2a)}{1\cdot 3}+\frac{\cos(4a)}{3\cdot 5}+\frac{\cos(6a)}{5\cdot 7}+\cdots \cdots  \)

Tenemos que hallar la función cuya serie de Fourier sea ésa.

- Vemos que es periódica de período \( \pi \) 
- Sabemos de antemano que denominadores del tipo \( \dfrac{1}{mk^2-n^2} \)  lo generan funciones trigonométricas.

- Calculamos entonces la SF de \( g(\theta)= a+b \sin \theta \)  en el intervalo \( [0,\pi] \)
  No incluyo un término \( c \cos \theta \)  pues generaría términos en senos (y solo tenemos cosenos)

\( \mathscr{F}(g(\theta))= a + \frac{2b}{\pi} - \frac{4b}{\pi}\displaystyle\sum_{k=1}^{\infty}\dfrac{\cos(2k)}{4k^2-1} \)
\( \;\;\Longrightarrow\;\;b=\frac{-\pi}{4}\quad,\quad a=\frac{1}{2} \)

\( \therefore\quad \frac{1}{2}-\frac{\pi}{4}\sin\theta \)  en \( [0,\pi] \)

[jacks]

Thanks Friends got it.