How close does #sum_(n=1)^oo 1/(n!), n="integers"# get to #e#?

Question

I was playing with sums of series and found that #sum_(n=1)^oo 1/(n!), n="integers"# got got to #e# on some devices I used, i.e. calculator or online.

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Answer 1 · 2017-07-04T18:43:30

See below.

It is easier to answer this question regarding #e^(-1)#

Considering the definition of #e^x# as

#e^x = sum_(k=0)^oo x^k/(k!)# we have

#e^(-x) = sum_(k=0)^oo(-1)^kx^k/(k!)# and then

#e^(-1) = sum_(k=0)^oo(-1)^k/(k!)#

This is a alternating convergent series such that #abs(a_(k+1)) < abs(a_k)#

then we know that the truncation error is less than the last non considered term.