I am given the series (e^n)/(3^(n-1))...Don't forget the Riemann sum symbol in front of this. It appears to be a geometric series. This would mean its ar^n

and a = 1/3 and r = e. I then do a bit of manipulation and get 1/3-1/3(e^n) and then Sn = (1-e^n)/(3-3e)

since r>1, the sequence {r^n} is divergent. Lim n--> infinity Sn does not exist. The geometric series is divergent.