Re: What is the error of the truncated exponential series?

["G. A. Edgar"]

In article <1159608550.179699.33700@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
<heinrich_neumaier@xxxxxxxxx> wrote:

> Summing from 0 to n the terms 1/n!, one gets an approximation for
> e=2.71828...
> 
> Is there a formula for the difference between the result  (as a
> function of n) and e?
> 
> Heinz
> 

A simple one to carry out is to estimate by a geometric series.


1/(n+1)! + 1/(n+2)! + 1/(n+3)! + ... <=
1/(n+1)! * [1 + 1/(n+2) + 1/(n+2)^2 + ... ] =
(n+2)/[(n+1)*(n+1)!]

-- 
G. A. Edgar                              http://www.math.ohio-state.edu/~edgar/