We consider the function $F : ] 0 , + \infty [ \rightarrow \mathbb { R }$ defined by $F ( x ) = \int _ { 1 } ^ { + \infty } e ^ { - t / x } t ^ { - 1 } d t$. For $N \in \mathbb { N } ^ { * }$ and $x > 0$, we set $$\begin{aligned}
S _ { N } ( x ) & = \sum _ { k = 1 } ^ { N } ( - 1 ) ^ { k - 1 } ( k - 1 ) ! x ^ { k } e ^ { - 1 / x } \\
R _ { N } ( x ) & = ( - 1 ) ^ { N } N ! x ^ { N } \int _ { 1 } ^ { + \infty } e ^ { - t / x } t ^ { - ( N + 1 ) } d t
\end{aligned}$$ The relative error is $E _ { N } ( x ) = \left| \frac { R _ { N } ( x ) } { F ( x ) } \right|$. Verify that $E _ { 4 } \left( \frac { 1 } { 10 } \right) \leqslant 3.10 ^ { - 3 }$.
We consider the function $F : ] 0 , + \infty [ \rightarrow \mathbb { R }$ defined by $F ( x ) = \int _ { 1 } ^ { + \infty } e ^ { - t / x } t ^ { - 1 } d t$.
For $N \in \mathbb { N } ^ { * }$ and $x > 0$, we set
$$\begin{aligned}
S _ { N } ( x ) & = \sum _ { k = 1 } ^ { N } ( - 1 ) ^ { k - 1 } ( k - 1 ) ! x ^ { k } e ^ { - 1 / x } \\
R _ { N } ( x ) & = ( - 1 ) ^ { N } N ! x ^ { N } \int _ { 1 } ^ { + \infty } e ^ { - t / x } t ^ { - ( N + 1 ) } d t
\end{aligned}$$
The relative error is $E _ { N } ( x ) = \left| \frac { R _ { N } ( x ) } { F ( x ) } \right|$.
Verify that $E _ { 4 } \left( \frac { 1 } { 10 } \right) \leqslant 3.10 ^ { - 3 }$.