Question 127883
I'm presuming that the series to which you refer is {{{e^x=sum((x^n)/n!,n=0,infinity)}}}


The first 5 terms:


{{{1+x+(x^2/2!)+(x^3/3!)+(x^4/4!)}}}


{{{1 + (-0.5) + (0.25/2) + (-0.125/6) + (0.0625/24)}}}


{{{1 - 0.5 + 0.125 - 0.0208 + 0.0026 = .6068}}} (roughly)


Calculator result: {{{0.6065}}}  Certainly close enough for government work.