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

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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.

I was playing with sums of series and found that

##### 1 Answer

Jul 4, 2017

See below.

#### Explanation:

It is easier to answer this question regarding

Considering the definition of

This is a alternating convergent series such that

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