Consider the integral I(k)=∫01xk(1−xk)dxI(k) = \int_0^1 x^k (1-x^k) dxI(k)=∫01xk(1−xk)dx for k>0k > 0k>0. If we define S=∑n=1∞I(n)S = \sum_{n=1}^{\infty} I(n)S=∑n=1∞I(n), what is the numerical value of SSS?
111
0.50.50.5
∞\infty∞
π26−1\frac{\pi^2}{6} - 16π2−1