# 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.

Jul 4, 2017

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 $\left\mid {a}_{k + 1} \right\mid < \left\mid {a}_{k} \right\mid$

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