$f(x)=\sqrt{1-\cos x}\[2ex]=\sqrt{2{\sin}^2\left(\frac{x}{2}\right)}\[2ex]=\sqrt{2}\sin{\left(\frac{x}{2}\right)}\[3ex] Let\ c=0\ and\ c+2l=2\pi{}\[2ex] \therefore{}0+2l=2\pi{}\[2ex] \therefore{}l=\pi{}\[2ex]Now,\[2ex] $ $\displaystyle a_0=\frac{1}{l}\int_c^{c+2l}f\left(x\right)dx\[2ex] \displaystyle =\frac{1}{\frac{\pi{}}{2}}\int_0^{\pi{}}\sqrt{2}\sin{\left(\frac{x}{2}\right)}dx\[2ex] \displaystyle =\frac{2\sqrt{2}}{\pi{}}{\left[\frac{-\cos{\left(\frac{x}{2}\right)}}{\frac{1}{2}}\right]}_0^{\pi{}}\[2ex] \displaystyle =\frac{-4\sqrt{2}}{\pi{}}\left[cos\left(\frac{x}{2}\right)-\cos0\right]\[2ex] \displaystyle =\frac{-4\sqrt{2}}{\pi{}}\left[0-1\right]\[2ex] \displaystyle =\frac{-4\sqrt{2}}{\pi{}}\[2ex] $ $\displaystyle a_n=\frac{1}{l}\int_c^{c+2l}f\left(x\right)\cos\left(\frac{n\pi{}x}{l}\right)dx\[2ex] \ \ \displaystyle =\frac{1}{\frac{\pi{}}{2}}\int_0^{\pi{}}\sqrt{2}\sin{\left(\frac{x}{2}\right)\cos\left(\frac{n\pi{}x}{\frac{\pi{}}{2}}\right)}dx\[2ex] \displaystyle =\frac{\sqrt{2}}{\pi{}}\int_0^{\pi{}}2\cos\left(2nx\right)\sin\left(\frac{x}{2}\right)dx\[2ex] \displaystyle =\frac{\sqrt{2}}{\pi{}}\int_0^{\pi{}}\left[\sin{\left(2nx+\frac{x}{2}\right)-\sin{\left(2nx-\frac{x}{2}\right)}}\right]dx\[2ex] \displaystyle =\frac{\sqrt{2}}{\pi{}}\int_0^{\pi{}}\left[\sin{\frac{\left(4n+1\right)x}{2}-\sin{\frac{\left(4n-1\right)x}{2}}}\right]dx\[2ex] \displaystyle =\frac{\sqrt{2}}{\pi{}}{\left[-\frac{\cos{\frac{\left(4n+1\right)x}{2}}}{\frac{\left(4n+1\right)}{2}}{+\frac{\cos{\frac{\left(4n-1\right)x}{2}}}{\frac{\left(4n-1\right)}{2}}}\right]}_0^{\pi{}}\[2ex] \displaystyle =\frac{\sqrt{2}}{\pi{}}\left{\left[-\frac{2\cos{\frac{\left(4n+1\right)\pi{}}{2}}}{\frac{\left(4n+1\right)}{2}}{+\frac{2\cos{\frac{\left(4n-1\right)\pi{}}{2}}}{\frac{\left(4n-1\right)}{2}}}\right]-\left[-\frac{2\cos0}{\left(4n+1\right)}{+\frac{2\cos0}{\left(4n-1\right)}}\right]\right}\[4ex] $ $Consider\[2ex]\cos{\frac{\left(4n\pm{}1\right)\pi{}}{2}=\cos\left(2n\pi{}\pm{}\frac{\pi{}}{2}\right)}\[2ex]=\cos\dfrac{\pi{}}{2}=0\[2ex] \therefore{} \displaystyle a_n=\frac{\sqrt{2}}{\pi{}}\left{0-0+\frac{2}{4n+1}-\frac{2}{4n-1}\right}\[2ex] \displaystyle =\frac{2\sqrt{2}}{\pi{}}\left{\frac{4n-1-4n-1}{\left(4n+1\right)\left(4n-1\right)}\right}\[2ex] \displaystyle =\frac{2\sqrt{2}}{\pi{}}\times{}\frac{-2}{16n^2-1}\[2ex]=\dfrac{-4\sqrt{2}}{\pi{}\left(16n^2-1\right)}\[2ex]

$ $Consider\[2ex] \displaystyle b_n=\frac{1}{l}\int_c^{c+2l}f\left(x\right)\sin\left(\frac{n\pi{}x}{l}\right)dx\[2ex] \displaystyle =\frac{1}{\frac{\pi{}}{2}}\int_0^{\pi{}}\sqrt{2}\sin{\left(\frac{x}{2}\right)\sin\left(\frac{n\pi{}x}{\frac{\pi{}}{2}}\right)}dx\[2ex] \displaystyle =\frac{\sqrt{2}}{\pi{}}\int_0^{\pi{}}2\sin \left(2nx\right)\sin\left(\frac{x}{2}\right)dx\[2ex] \displaystyle =\frac{\sqrt{2}}{\pi{}}\int_0^{\pi{}}\left[\cos{\left(2nx+\frac{x}{2}\right)-\cos{\left(2nx-\frac{x}{2}\right)}}\right]dx\[2ex] \displaystyle =\frac{\sqrt{2}}{\pi{}}\int_0^{\pi{}}\left[\cos{\frac{\left(4n+1\right)x}{2}-\cos{\frac{\left(4n-1\right)x}{2}}}\right]dx\[2ex] \displaystyle =\frac{\sqrt{2}}{\pi{}}{\left[-\frac{\sin{\frac{\left(4n+1\right)x}{2}}}{\frac{\left(4n+1\right)}{2}}{+\frac{\sin{\frac{\left(4n-1\right)x}{2}}}{\frac{\left(4n-1\right)}{2}}}\right]}_0^{\pi{}}\[2ex] \displaystyle =\frac{\sqrt{2}}{\pi{}}\left{\left[-\frac{2\sin{\frac{\left(4n+1\right)\pi{}}{2}}}{\frac{\left(4n+1\right)}{2}}{+\frac{2\sin{\frac{\left(4n-1\right)\pi{}}{2}}}{\frac{\left(4n-1\right)}{2}}}\right]-\left[-\frac{2\sin0}{\left(4n+1\right)}{+\frac{2\sin0}{\left(4n-1\right)}}\right]\right}\[2ex] $ $Consider\[2ex] \displaystyle \sin{\frac{\left(4n\pm{}1\right)\pi{}}{2}=\sin\left(2n\pi{}\pm{}\frac{\pi{}}{2}\right)}\[2ex] \displaystyle =\pm{}\sin\frac{\pi{}}{2}=\pm{}1\[2ex] $ $\therefore{}\ \displaystyle b_n=\frac{\sqrt{2}}{\pi{}}\left{\left[\frac{-1}{4n+1}-\frac{1}{4n-1}\right]-\left[0+0\right]\right}\[2ex] \displaystyle =\frac{\sqrt{2}}{\pi{}}\left{\frac{-4n-1-4n+1}{\left(4n+1\right)\left(4n-1\right)}\right}\[2ex] \displaystyle =\frac{\sqrt{2}}{\pi{}}\times{}\frac{-8n}{16n^2-1}\[2ex] \displaystyle =\frac{-8n\sqrt{2}}{\pi{}\left(16n^2-1\right)} $   $In\ Fourier\ series,\[3ex] \displaystyle f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty{}}an\cos\left(\frac{n\pi{}x}{l}\right)+\sum_{n=1}^{\infty{}}b_n\sin\left(\frac{n\pi{}x}{l}\right)\[2ex]\sqrt{1-\cos x} =\frac{2\sqrt{2}}{\pi{}}+\sum_{n=1}^{\infty{}}\frac{-4\sqrt{2}}{\pi{}\left(16n^2-1\right)}\cos\left(\frac{n\pi{}x}{\frac{\pi{}}{2}}\right)+\sum_{n=1}^{\infty{}}\frac{-8n\sqrt{2}}{\pi{}\left(16n^2-1\right)}\sin\left(\frac{n\pi{}x}{\frac{\pi{}}{2}}\right)\[2ex] \displaystyle =\frac{2\sqrt{2}}{\pi{}}-\frac{4\sqrt{2}}{\pi{}}\sum_{n=1}^{\infty{}}\frac{\cos2nx}{\left(16n^2-1\right)}-\frac{8\sqrt{2}}{\pi{}}\sum_{n=1}^{\infty{}}\frac{n\sin2nx}{\left(16n^2-1\right)}\ \rightarrow{}(i)

$ $\displaystyle =\frac{2\sqrt{2}}{\pi{}}-\frac{4\sqrt{2}}{\pi{}}\left[\frac{\cos2x}{\left(16\bullet{}1^2-1\right)}+\frac{\cos4x}{\left(16{\bullet{}2}^2-1\right)}+…\right]-\frac{8\sqrt{2}}{\pi{}}\left[\frac{\sin2x}{\left(16{\bullet{}1}^2-1\right)}+\frac{2\sin 2nx}{\left(16\bullet{}2^2-1\right)}+…\right] \[2ex] $ $Deduction:\[2ex] Put\ x=0\ in\[2ex] \displaystyle (i)\therefore{}\sqrt{1-cos0}=\frac{2\sqrt{2}}{\pi{}}-\frac{4\sqrt{2}}{\pi{}}\sum_{n=1}^{\infty{}}\frac{cos0}{\left(16n^2-1\right)}\[2ex]\displaystyle \therefore{}0-\frac{2\sqrt{2}}{\pi{}}=-\frac{4\sqrt{2}}{\pi{}}\sum_{n=1}^{\infty{}}\frac{1}{\left(16n^2-1\right)}\[2ex] \displaystyle \therefore{}\frac{1}{2}=\sum_{n=1}^{\infty{}}\frac{1}{{\left(4n\right)}^2-1}$