maths probability?

Let P(k) = C/ k^4 , where C is a constant and C > 0
( Why C > 0 ? Because C/ k^4 is a probability and k > 0 )

P(Odd)
= P(1) + P(3) + P(5) + ..... + P(2m-1)
= C/ 1^4 + C/ 3^4 + C/ 5^4 + ..... + C/ (2m-1)^4
= C[ 1/ 1^4 + 1/ 3^4 + 1/ 5^4 + ..... + 1/ (2m-1)^4 ]

P(Even)
= P(2) + P(4) + P(6) + ..... + P(2m)
= C[ 1/ 2^4 + 1/ 4^4 + 1/ 6^4 + ..... + 1/ (2m)^4 ]

1/ 1^4 > 1/ 2^4
1/ 3^4 > 1/ 4^4
1/ 5^4 > 1/ 6^4
.....................
1/ (2m-1)^4 > 1/ (2m)^4

Thus, P(Odd) > P(Even)
2*P(Odd) = P(Odd) + P(Odd) > P(Odd) + P(Even) = 1
P(Odd) > 1/2

Ans: C