Maths(infinite sequence)

let (-1)^(n-1)*(1/n) be the nth term of the sequence
the common different would be
(-1)^(n-1)*(1/n) - (-1)^(n)*(1/(n+1))
=(-1)^(n-1)*(1/n + 1/(n+1))

Let nth term be (-1)^(n+1)．(1/(2n-1))

therefore the common different is

T(n+1)-T(n) = (-1)^(n+2)．(1/(2n+1)) + (-1)^(n+1)．(1/(2n-1)
=(-1)^(n+1)[ 1/(2n-1) - 1/(2n+1))
=(-1)^(n+1)[2/(4n^2-1)]