(B) Recall the Maclaurin series of \(e^{x}\) :
\[
e^{x}=\sum_{n=0}^{\infty} \frac{x^{n}}{n!}
\]
with radius of convergence \(r=\infty\).
You replace \(x\) by \(-x\) in the Maclaurin series and simplify, yielding
\[
\begin{aligned}
\frac{e^{-x}-1}{x} & =\frac{1}{x}\left(-1+\sum_{n=0}^{\infty} \frac{(-x)^{n}}{n!}\right)=\frac{1}{x}\left(-1+\sum_{n=0}^{\infty} \frac{(-1)^{n} x^{n}}{n!}\right)=\frac{1}{x}\left(-1+1+\sum_{n=1}^{\infty} \frac{(-1)^{n} x^{n}}{n!}\right) \\
& =\frac{1}{x}\left(\sum_{n=1}^{\infty} \frac{(-1)^{n} x^{n}}{n!}\right)=\sum_{n=1}^{\infty} \frac{(-1)^{n} x^{n-1}}{n!}
\end{aligned}
\]

Next, you integrate to obtain
\[
\begin{aligned}
\int_{0}^{1}\left(\sum_{n=1}^{\infty} \frac{(-1)^{n} x^{n-1}}{n!}\right) d x & =\sum_{n=1}^{\infty}\left(\frac{(-1)^{n}}{n!} \int_{0}^{1} x^{n-1} d x\right)=\sum_{n=1}^{\infty}\left(\frac{(-1)^{n}}{n!}\left[\frac{x^{n}}{n}\right]_{0}^{1}\right]_{n=1}^{\infty}\left(\frac{(-1)^{n}}{n!}\left[\frac{1}{n}-0\right]\right)=\sum_{n=1}^{\infty}\left(\frac{(-1)^{n}}{n \cdot n!}\right)=\sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{(n+1) \cdot(n+1)!}
\end{aligned}
\]

This is an alternating series, which you will denote by \(s\). Let \(b_{n}=\frac{1}{(n+1) \cdot(n+1)!}\), and denote by \(s_{n}\) the \(n\)th partial sum. When estimating alternating series, you use \(\left|s-s_{n}\right|< b_{n+1}\). Hence,
\[
\left|s-s_{n}\right| \leq \frac{1}{(n+2) \cdot(n+2)!}
\]

Observe that when \(n=5,\left|s-s_{5}\right| \leq \frac{1}{7 \cdot 7!} \approx 0.00002\). Computing
\[
s_{5}=-1+\frac{1}{2 \cdot 2!}-\frac{1}{3 \cdot 3!}+\frac{1}{4 \cdot 4!}-\frac{1}{5 \cdot 5!}+\frac{1}{6 \cdot 6!} \approx-0.7965
\]

Observe that an error of 0.00002 does not affect the third place after the decimal, so your approximation -0.796 is correct to three decimal places.

Use Maclaurin series to approximate the integral $$\(\int_{0}^{1} \frac{e^{-x}-1}{x} d x\)$$ to three

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