Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics V / 4 / c

At $x=0$, the Fourier sum is $f(0)=1$, and only the odd cosine coefficients contribute. Therefore $1=1/4+(6/{\pi }^2){\sum }_{k\ge 0}(2k+1{)}^{-2}$, givingAt $x=1/2$, the sum is $f(1/2)=1/2$. All nonzero cosine terms vanish, and the sine terms have alternating signs over odd indices. Thus $1/2=1/4+(1/\pi ){\sum }_{k\ge 0}(-1{)}^k/(2k+1)$, soThis is the Leibniz formula for π obtained from the same Fourier series.