Sum

_{n=2}

^{infinity}(1/(n(n-1))

=Sum

_{n=2}

^{infinity}(1/(n^2)+1/(n^3)+1/(n^4)+...)

=Sum

_{n=2}

^{infinity}(Sum

_{i=2}

^{infinity}(1/(n^i)))

=Sum

_{i=2}

^{infinity}(Sum

_{n=2}

^{infinity}(1/(n^i)))

=Sum

_{i=2}

^{infinity}(Z(i)-1), where Z is the Riemann Zeta Function.

But Sum

_{n=2}

^{x}(1/(n(n-1))=(x-1)/x

So Sum

_{n=2}

^{infinity}(1/(n(n-1))=1

Therefore Sum

_{i=2}

^{infinity}(Z(i)-1)=1

Is this right?