Hello,

I haven't checked your formulae, it looks long

However, note that the power series of the exponential starts atn=0, and the power series of the logarithm stars atn=1.

This should solve the problem of dividing by 0.

The MacLaurin series is a power series. The expansion of a function into a series is unique. So expanding a function as a power series is equivalent to getting the MacLaurin series. (I don't know if I'm clear here ? :s)

To simplify calculations, you can use the Cauchy product for series. But again, be careful that the series should start at the same point