Find half range cosine series of f(x)= sinx in 0 ≤ x ≤ π Hence deduce that [\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+?=\frac{1}{2}]

$\displaystyle For\ half\ range\ cosine\ ,\ b_n=0\ \\[2ex] \displaystyle f\left(x\right)=\sin x\ ,\ l=\pi{}\ \\[2ex] \displaystyle a_0=\frac{2}{l}\ \int_0^lf(x)dx \\[2ex] \displaystyle a_0=\frac{2}{\pi{}}\ \int_0^{\pi{}}\sin{x\ }\ dx\ \\[2ex]a_0=\frac{2}{\pi{}}{\left[-\cos x\right]}_0^{\pi{}}\\[2ex] a_0=-\dfrac{2}{\pi{}}\left[-1-1\right] \\[2ex] \displaystyle a_0=\frac{4}{\pi{}} \\[2ex] $

$\displaystyle a_n=\frac{2}{l}\ \int_0^lf(x)\cos{\left(\frac{n\pi{}x}{l}\right)}dx \\[2ex] \displaystyle a_n=\frac{2}{\pi{}}\ \int_0^{\pi{}}\sin x\cos{nx}dx=\frac{1}{\pi{}}\int_0^{\pi{}}\sin{\left(n+1\right)x}+\sin{\left(1-n\right)x}\ dx \\[2ex] \displaystyle =\frac{-1}{\pi{}}{\left[\frac{\cos{\left(n+1\right)x}}{n+1}+\frac{\cos{\left(1-n\right)x}}{1-n}\right]}_0^{\pi{}}\ \ \ \ \ \ \ \ \ n\not=1\ \\[2ex] \displaystyle =\ -\frac{1}{\pi{}}\ \left[\frac{{\left(-1\right)}^{n+1}}{n+1}-\frac{{\left(-1\right)}^{n-1}}{n-1}+\frac{1}{n+1}-\frac{1}{n-1}\right] \\[2ex] \displaystyle =-\frac{1}{\pi{}}\left[\frac{{\left(-1\right)}^n2}{n^2-1}+\frac{2}{n^2-1}\right] \\[2ex] \displaystyle a_n=\frac{2}{\pi{}(1-n^2)}\left[1+{\left(-1\right)}^n\right]\ \ \ \ \\[2ex] \displaystyle for\ n=1,\ \\[2ex] $

$ \displaystyle a_1=\frac{2}{\pi{}}\int_0^{\pi{}}\sin x\cos xdx\\[2ex] \displaystyle a_1=\frac{1}{\pi{}}\int_0^{\pi{}}\sin{2x}dx\\[2ex] a_1=\ -\dfrac{1}{2\pi{}}\ {\left[\cos{2x}\right]}_0^{\pi{}}\\[2ex] a_1=\ -\dfrac{1}{2\pi{}}\left[1-1\right] \\[2ex] \displaystyle a_1=0 \\[2ex] \displaystyle \ \ \therefore{}Half\ range\ fourier\ cosine\ series\ is \\[2ex] \displaystyle f\left(x\right)=\frac{a_0}{2}+\sum_{n=1}^{\infty{}}a_n\ \cos{\frac{n\pi{}x}{l}}\ \\[2ex] \displaystyle \sin x=\frac{2}{\pi{}}+\sum_2^{\infty{}}\frac{2}{\pi{}\left(1-n^2\right)}\left[1+{\left(-1\right)}^n\right]\cos{nx}\ \\[2ex] \displaystyle put\ x=0, \\[2ex] \displaystyle 0=\frac{2}{\pi{}}-\frac{2}{\pi{}}\ \sum_2^{\infty{}}\frac{\left[1+{\left(-1\right)}^n\right]}{\left(n^2-1\right)}\ \\[2ex] \displaystyle 1=\frac{2}{3}+0+\frac{2}{15}+0+\frac{2}{35}+… \\[2ex] \displaystyle \therefore{},\ \frac{1}{2}=\frac{1}{1.3}\ +\frac{1}{3.5}+\frac{1}{5.7}+… \\[2ex] $