[net.math] equality for all

jhillis (01/25/83)

As was pointed out last time this was argued:

      .9999...  =   .9+.09+.009+.0009+...
                =   sum (.9)*(.01)**n        where n=0,1,2,3...
                =   (.9)*[1/(1-.01)]
                =   (.9)*(1/.9)
                =   1