How close does $\infty \sum n = 1 \frac{1}{n!}, n = integers$ $sum_(n=1)^oo 1/(n!), n="integers"$ get to $e$ $e$ ?

Question

I was playing with sums of series and found that $\infty \sum n = 1 \frac{1}{n!}, n = integers$ $sum_(n=1)^oo 1/(n!), n="integers"$ got got to $e$ $e$ 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 $e^{- 1}$ $e^(-1)$

Considering the definition of $e^{x}$ $e^x$ as

$e^{x} = \infty \sum k = 0 \frac{x^{k}}{k!}$ $e^x = sum_(k=0)^oo x^k/(k!)$ we have

$e^{- x} = \infty \sum k = 0 {(- 1)}^{k} \frac{x^{k}}{k!}$ $e^(-x) = sum_(k=0)^oo(-1)^kx^k/(k!)$ and then

$e^{- 1} = \infty \sum k = 0 \frac{{(- 1)}^{k}}{k!}$ $e^(-1) = sum_(k=0)^oo(-1)^k/(k!)$

This is a alternating convergent series such that $| a_{k + 1} | < | a_{k} |$ $abs(a_(k+1)) < abs(a_k)$

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