Verified answer

Let

f(x) = Σ [n=1,∞][ (2/(nπ))⋅sin(nπx) ].

Then

f(1/2)

= Σ [ n=1,∞ ][ (2/(nπ))⋅sin(nπ/2) ]

= (2/π)⋅Σ [ n=1,∞ ][ (1/n)⋅sin(nπ/2) ]

= (2/π)⋅Σ [ n=0,∞ ][ (1/(2n+1))⋅sin((2n+1)π/2) + (1/(2n))⋅sin((2n)π/2) ]

= (2/π)⋅Σ [ n=0,∞ ][ (1/(2n+1))⋅(-1)^n ].

Hence

Σ [ n=0,∞ ][ (1/(2n+1))⋅(-1)^n ] = (π/2)⋅f(1/2).