if 1÷1^2+1÷2^2+1÷3^2_____infinity=[(pie)^2]÷6 then value of 1÷1^2+1÷3^2+1÷5^2_____infinity is???

1÷1^2+1÷3^2+1÷5^2_____infinity=[(pie)^2]÷6 - {1÷2^2+1÷4^2+1÷6^2_____infinity}...(A) now {1÷2^2+1÷4^2+1÷6^2_____infinity}= summation of 1/((2n)^2)[ nth term of given sequence] up to infinity now1/((2n)^2)=1/(4xn^2)=(1/4)x(1/(n^2))...(1) we know 1÷1^2+1÷2^2+1÷3^2_____infinity=[(pie)^2]÷6 or summation of (1/(n^2) [ nth term of given sequence] up to infinity =[(pie)^2]÷6 hence (1)= (1/4)x([(pie)^2]÷6)=pie square by twenty four ..putting this in (A) 1÷1^2+1÷3^2+1÷5^2_____infinity=[(pie)^2]÷6 - {1÷2^2+1÷4^2+1÷6^2_____infinity}= [(pie)^2]÷6 - pie square by twenty four= pie square by eight .. if u are satisfied by my answer plz buddy can u clear my doubt ..its in thermal physics section with title doubt and i posted it on 20 th feb ..thnx