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

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.

1 Answer
Jul 4, 2017

Answer:

See below.

Explanation:

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.