<s>[INST] Approximate $e$ with $4$-digit chopping and $5$-degree Maclaurin-Taylor polynomial 
Number $e$ is defined as 
  $$ e=\sum^\infty_{n=0} \frac{1}{n!}.$$
  Use four-digit chopping arithmatic to compute the following 
  
  
*
  
*Approximations to $e$
$$ e\approx \sum^5_{n=0} \frac{1}{n!}$$


Attempt:
\begin{align}
\sum^5_{n=0} \frac{1}{n!}
&= \frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!} \\ &= \frac{1}{1}+\frac{1}{2}+\frac{1}{3 \cdot 2 \cdot 1}+\frac{1}{4 \cdot 3 \cdot 2 \cdot 1}+\frac{1}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} \\ 
&= 1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{1}{120}
\end{align}
to decimals
$$=1+.5000+0.16666666666+.04166666667+.0083333333333333$$
and I am not sure how $4$-digit chopping works.
I think that doing $4$-digit chopping is 
$$=1+.5000+0.1666+.04166+.008333=1.716593=1.716$$
I am getting that it is wrong according to webwork wich would never ever be wrong.

 [/INST] That is quite easy: you forgot the $1/ 0!$ in your calculations. Then you get something like $\mathrm e \approx2.716$. Also rounding in the end of your calculation might be the best.
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