Using recursion this way, you Sir is a master!

Super clever!!! Never come up with the idea

very smart, good code!

# 1/n! * (n! + (n-1)! + (n-2)! ......1!)      # can be factored down to:

Very clever

# n!/n! + (n-1)!/n! + (n-2)!/n! .....1!/n!    # write out the factorial e.g. 5 x 4 x 3 ... and you'll see it better

#  s(n) =  1   +    1/n    +  1/n(n-1) ...1/n!       # 1/n plays the largest role. In fact, after n=15, 
                                              # the 1st 6 decimals aren't even affected - trial and error
# s(n-1) = 1 + 1/(n - 1) + ... + 1/(n-1)!

# A recursive formula indicates s(n-1)/n + 1 = 1 + 1/n + 1/(n-1)n + ... + 1/n! =s(n)

good

That was the kind of solution I was looking for. I knew there would be "itertoolsless" solution using recursion but I couldnt solve it myself. Thanks! As a beginner getting mindblown by recursion this was really helpfull for me

to truncate to 6 decimal places

This could be improved: right now, you are calling b.lower() for every letter in a, and you are calling a.lower() for every letter in b.