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\problem{Problem 24, section 1.62} 

Let $\displaystyle b_{n} = \left(1+\frac{1}{n}\right)^{n}$ and let
$a_{p} = \displaystyle \sum_{k=0}^{p} \frac{1}{k!}$.  From Example 6
in the book, we know that the sequence $\{b_{n}\}$ is increasing and
bounded above and that its limit is $e$; therefore $b_{n}<e$ for all
$n$.  Also from Example 6 (in particular, about a third of the way
down on page 64), we know that $b_{n} < a_{n}$ for all $n$.

Therefore, we can follow the book's hint and try to conclude that
$\lim b_{n} \leq \lim a_{n}$ -- I don't think this is obvious, but it
can be justified using the definition of limit.\footnote{There are two
possibilities: either $a<e$ or $a \geq e$.  We want to rule out the
first of these, so let's use contradiction: assume that $a<e$.  Let
$\epsilon = (e-a)/2$, so that $e=a+2\epsilon$.  Since $\lim b_{n} =
e$, there is an $N$ so that $|e-b_{n}|<\epsilon$ for all $n \geq N$.
Since $b_{n}<e$ for each $n$, we can rewrite this as
$e-b_{n}<\epsilon$, or $e-\epsilon<b_{n}$.  Therefore $a+
\epsilon<b_{n}<a_{n}$.  This means that $|a_{n}-a|>\epsilon$.  Since
$a_{n}$ is farther than $\epsilon$ from $a$ for all $n \geq N$, the
limit $\lim a_{n}$ cannot equal $a$.  This is a contradiction;
therefore $a \geq e$.}  More importantly, though, we \emph{don't
actually need this} for the rest of the problem: all we need is that
$b_{n} < a_{n}$ for all $n$.

Equation (1.62-3) says that 
\[
b_{n} = 1 + \frac{1}{1!} + \frac{1-\frac{1}{n}}{2!} + \dotsb + 
\frac{\left(1-\frac{1}{n}\right) \dotsb \left(1-\frac{n-1}{n}\right)}{n!}.
\]
This is a sum with $n+1$ terms (terms with denominators $0!$, $1!$,
$2!$, \dots, $n!$).  We could write it as
\[
b_{n} = 1 + \sum_{k=1}^{n} \frac{\left(1-\frac{1}{n}\right) \dotsb
\left(1-\frac{k-1}{n}\right)}{k!}.
\]
For any $p$ with $1<p<n$, if we take the first $p+1$ terms, then the
sum is smaller than $b_{n}$, and that sum is
\[
1 + \frac{1}{1!} + \frac{1-\frac{1}{n}}{2!} + \dotsb + 
\frac{\left(1-\frac{1}{n}\right) \dotsb
\left(1-\frac{p-1}{n}\right)}{p!}
= 1 + \sum_{k=1}^{p} \frac{\left(1-\frac{1}{n}\right) \dotsb
\left(1-\frac{k-1}{n}\right)}{k!}.
\]
Denote this sum by $c_{p,n}$.
Fix $p$ and let $n \rightarrow \infty$ here: the denominator of each
fraction goes to $1$, so the 
limit\footnote{
We can break this step down: for each positive integer $i$, the
expression $1-i/n$ goes to $1$ as $n \rightarrow \infty$.  Since the
limit of a product is the product of the limits, then for each $k$,
\[
\lim_{n \rightarrow \infty} \left(1-\frac{1}{n}\right) \dotsb
\left(1-\frac{k-1}{n}\right) = 1,
\]
and since the limit of a quotient is the quotient of the limits, 
\[
\lim_{n \rightarrow \infty}
\frac{\left(1-\frac{1}{n}\right) \dotsb
\left(1-\frac{k-1}{n}\right)}{k!}  = \frac{1}{k!}.
\]
Finally, since the limit of a sum is the sum of the limits, we have
\[
\lim_{n \rightarrow \infty} \left(1 + \sum_{k=1}^{p}
\frac{\left(1-\frac{1}{n}\right) \dotsb
\left(1-\frac{k-1}{n}\right)}{k!} \right) = 1 + \sum_{k=1}^{p}
\frac{1}{k!} = a_{p}.
\]
}
is
\[
\lim_{n \rightarrow \infty} c_{p,n} = 1 + \frac{1}{1!} + \frac{1}{2!}
+ \dotsb + \frac{1}{p!} = a_{p}.
\]
Since $c_{p,n} < b_{n}$ for all $p$ and $n$, and since $b_{n} < e$ for
all $n$, we have that $c_{p,n} < e$ for all $p$ and $n$.  Therefore
the analogue of Theorem XI for sequences says that $\lim c_{p,n} \leq
e$, which means that $a_{p} \leq e$.  Therefore we have (replacing $p$
with $n$):
\[
b_{n} < a_{n} \leq e \quad \text{for all $n$}.
\]
Now apply the squeeze principle: since $\lim b_{n} = e$, we conclude
that $a = e$, as desired.

Alternatively, once we know that $a_{p} \leq e$ for all $p$, then we
apply the analogue of Theorem XI for sequences again to conclude that
$a \leq e$.  If we have also established that $a \geq e$, then these
together imply that $a=e$.

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