<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\section{Defining $e$}

\problem{}
Recall and prove the binomial theorem.

\begin{solution}
	The binomial theorem:
	$$(a + b)^n = \sum_{i=0}^{n} \binom{n}{i}a^ib^{n-i}$$

	We usually prove this by induction.
\end{solution}

\vfill




\problem{}&lt;e_n&gt;
Prove the following:

$$ \bigg(1 + \frac{1}{n} \bigg)^n &lt; 3 - \frac{1}{n}\ \text{for all } n = 3, 4, 5, ...$$

This proves that that the sequence $e_n = (1 + \frac{1}{n})^n$, $n = 1, 2, ...$ is bounded from above, \par
since $e_n &lt; 3\ \forall n \in \mathbb{N}$.

\begin{solution}
	$$
		\bigg(1 + \frac{1}{n}\bigg)^n =
		\sum_{k=0}^n \binom{n}{k}\frac{1}{n^i} =
		2 + \sum_{k=2}^n \binom{n}{k}\frac{1}{n^i} =
		2 + \sum_{k=2}^n \frac{1}{k!}\frac{(n)(n-1)...(n-k+1)}{n^k} &lt;
		2 + \sum_{k=2}^n \frac{1}{k!}
	$$

	$$
		2 + \sum_{k=2}^n \frac{1}{k!} &lt;
		2 + \sum_{k=2}^n \frac{1}{k(k-1)} =
		2 + 1 - \frac{1}{n} = 3 - \frac{1}{n}
	$$
\end{solution}

\vfill
\pagebreak



\theorem{Bernoulli's inequality}&lt;bernoulli&gt;
$(x + 1)^n \geq 1 + nx$ for $x \geq -1$ and $n \in \mathbb{N}$

\problem{}
Use induction to prove \ref{bernoulli}

\vfill



\problem{}&lt;e_n_inc&gt;
Use \ref{bernoulli} to prove that the $e_n$ defined in \ref{e_n} is monotonically increasing.

\begin{solution}
	We want to show that the following is true:
		$$
			\bigg(
				1 + \frac{1}{n + 1}
			\bigg)^{n+1}
			&gt;
			\bigg(
				1 + \frac{1}{n}
			\bigg)^n
		$$

	This inequality is equivalent to
		$$
			\Bigg(
				\frac{
					1 + \frac{1}{n+1}
				}{
					1 + \frac{1}{n}
				}
			\Bigg)^{n+1}
			&gt;
			\bigg(
				1 + \frac{1}{n}
			\bigg)^{-1}
			= \frac{1}{1 + \frac{1}{n}}
			= \frac{n}{n+1}
			= 1 - \frac{1}{n+1}
		$$

	Also,
		$$
			\frac{
				1 + \frac{1}{n+1}
			} {
				1 + \frac{1}{n}
			}
			= 1 - \frac{1}{(n + 1)^2}
		$$

	\ref{bernoulli} tells us that
		$$
			\bigg(
				1 - \frac{1}{(n+1)^2}
			\bigg) ^ {n+1}
			= 1 - \frac{n+1}{(n+1)^2}
			= 1 - \frac{1}{n+1}
		$$

	Since this is equivalent to our original inequality, we are done.

\end{solution}

\vfill
\pagebreak



\definition{}
\ref{e_n}, \ref{e_n_inc}, and \ref{limexists} tell us that $e_n$ has a limit. \par
Let us define $e$:
$$e = \lim_{n\to\infty}{e_n} = \lim_{n\to\infty}{\bigg(1 + \frac{1}{n} \bigg)^n}$$

\vfill


\problem{}
Derive the formula for $V(t)$ if the annual rate $r$ is compounded continuously.

\begin{solution}
	For a rate $r$ compounded $n$ times per year, we have $V(t) = P(1 + \frac{r}{n})^{nt}$.

	$\lim_{n\to\infty}{(V(t))} = \lim_{n\to\infty}{P(1 + \frac{r}{n})^{nt}}$

	Let $a = \frac{n}{r}$ \par

	$\lim_{n\to\infty}{a} = \infty$, so $\lim_{n\to\infty}{V(t)} = \lim_{a\to\infty}{V(t)}$ \par

	Substituting $a$ for $n$, we get \par
	$P(1 + \frac{r}{n})^{nt} = P(1 + \frac{1}{a})^{art}$ \par

	And finally, we can evaluate \par
	$\lim_{a\to\infty}{P (1 +  \frac{1}{a})^{art}} = Pe^{rt}$
\end{solution}

\vfill
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