Answer: (a) \(\displaystyle \int_0^1 e^{1-x}\,dx\lt U_n\), where \(\displaystyle U_n=\frac{e-1}{n\left(1-e^{-1/n}\right)}\).

(b) \(\displaystyle L_n=\frac{e-1}{n\left(e^{1/n}-1\right)}\).

(c) \(\displaystyle U_n-L_n=\frac{e-1}{n}\), so \(\displaystyle \lim_{n\to\infty}(U_n-L_n)=0\).

(d) \(\displaystyle z\left(1-e^{-1/z}\right)=1-\frac{1}{2z}+\frac{1}{6z^2}+\cdots\), hence \(\displaystyle \lim_{n\to\infty}U_n=e-1\).

(a) Since \(y=e^{1-x}\) is decreasing on \([0,1]\), the left-endpoint rectangles give an upper bound.

Each rectangle has width \(\frac1n\), so

\(\displaystyle \int_0^1 e^{1-x}\,dx\lt \frac1n\left(e^{1-0/n}+e^{1-1/n}+\cdots+e^{1-(n-1)/n}\right).\)

Thus

\(\displaystyle \int_0^1 e^{1-x}\,dx\lt \frac{e}{n}\sum_{r=0}^{n-1}e^{-r/n}.\)

This is a geometric series with ratio \(e^{-1/n}\), so

\(\displaystyle \sum_{r=0}^{n-1}e^{-r/n}=\frac{1-e^{-1}}{1-e^{-1/n}}.\)

Hence

\(\displaystyle \int_0^1 e^{1-x}\,dx\lt \frac{e}{n}\cdot\frac{1-e^{-1}}{1-e^{-1/n}}=\frac{e-1}{n\left(1-e^{-1/n}\right)}=U_n.\)

(b) Using right-endpoint rectangles gives a lower bound, so

\(\displaystyle \int_0^1 e^{1-x}\,dx\gt \frac1n\left(e^{1-1/n}+e^{1-2/n}+\cdots+e^{1-n/n}\right).\)

This is

\(\displaystyle \frac{e}{n}\sum_{r=1}^{n}e^{-r/n}.\)

Summing the geometric series,

\(\displaystyle \sum_{r=1}^{n}e^{-r/n}=\frac{e^{-1/n}(1-e^{-1})}{1-e^{-1/n}}.\)

So

\(\displaystyle L_n=\frac{e}{n}\cdot\frac{e^{-1/n}(1-e^{-1})}{1-e^{-1/n}}=\frac{e-1}{n}\cdot\frac{e^{-1/n}}{1-e^{-1/n}}=\frac{e-1}{n\left(e^{1/n}-1\right)}.\)

(c) Now

\(\displaystyle U_n-L_n=\frac{e-1}{n\left(1-e^{-1/n}\right)}-\frac{e-1}{n\left(e^{1/n}-1\right)}.\)

Since \(\displaystyle \frac{1}{1-e^{-1/n}}=\frac{e^{1/n}}{e^{1/n}-1}\),

\(\displaystyle U_n-L_n=\frac{e-1}{n}\left(\frac{e^{1/n}}{e^{1/n}-1}-\frac{1}{e^{1/n}-1}\right)=\frac{e-1}{n}.\)

Therefore

\(\displaystyle \lim_{n\to\infty}(U_n-L_n)=\lim_{n\to\infty}\frac{e-1}{n}=0.\)

(d) Using the Maclaurin series \(\displaystyle e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\), put \(x=-\frac1z\):

\(\displaystyle e^{-1/z}=1-\frac1z+\frac{1}{2z^2}-\frac{1}{6z^3}+\cdots.\)

Hence

\(\displaystyle z\left(1-e^{-1/z}\right)=z\left(1-\left(1-\frac1z+\frac{1}{2z^2}-\frac{1}{6z^3}+\cdots\right)\right)=1-\frac{1}{2z}+\frac{1}{6z^2}+\cdots.\)

Now

\(\displaystyle U_n=\frac{e-1}{n\left(1-e^{-1/n}\right)}=\frac{e-1}{n(1-e^{-1/n})}.\)

From the expansion with \(z=n\),

\(\displaystyle n(1-e^{-1/n})\to 1\quad \text{as } n\to\infty.\)

So

\(\displaystyle \lim_{n\to\infty}U_n=e-1.\)