Calculus- series problem (show work)?

Question

Approximate the sum of the series correct to four decimal places.
((-1)^n)/((3^n)*n!)

Answer 1

((-1)^n)/((3^n)*n!) = (-1/3)^n / n!

The Taylor series expansion for e^(-1/3) is:

1 - (1/3)/1! + (1/9)/2! - (1/27)/3! + ...

So the sum of the series should be e^(-1/3)
which is 0.7165 to four decimal places.

Answer 2

In an alternate series the error is less than the first term left out
(taken in abs.value)
As we make errors by cutting decimals.Ex 1/3=0.3333
let´s try 1/3^n*n!<1/10^5 we have to take n=6 and sum five terms of the series
Starting with n=1 (not shown)
-1/3+1/18 -1/162+1/1944 -1/29160=-0.2835
If you have to start with n=0 sum 1 to the former