Question

How do you use a power series to find the exact value of the sum of the series `pi/4-(pi/4)^3/(3!)+(pi/4)^5/(5!)-(pi/4)^7/(7!) + …` ?

Answer 1

Since

`x-x^3/{3!}+x^5/{5!}-x^7/{7!}+cdots=sinx`,

`pi/4-(pi/4)^3/{3!}+(pi/4)^5/{5!}-(pi/4)^7/{7!}+cdots=sin(pi/4)=1/sqrt{2}`