05A10-ProofOfUpperAndLowerBoundsToBinomialCoefficient.tex

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\pmtitle{proof of upper and lower bounds to binomial coefficient}
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\begin{document}
Let $2 \le k \le n$ be natural numbers. We'll first prove the
inequality
\begin{equation*}
{n \choose k} \le \left(\frac{ne}{k}\right)^k.
\end{equation*}
We rewrite ${n \choose k}$ as
\begin{eqnarray*}
{n \choose k} &=& \frac{n (n-1)\cdots (n-k+1)}{k!} \\
&=& \left(1 -\frac{1}{n}\right)\cdots \left(1
-\frac{k-1}{n}\right)\cdot \frac{n^k}{k!}
\end{eqnarray*}
Since each of the parenthesized factors lies between $0$ and $1$, we have
\begin{equation*}
{n \choose k} < \frac{n^k}{k!}
\end{equation*}
Since all the terms of the series $e^k = \sum_{n=0}^{\infty} k^n / n!$ are positive when $k$ is a positive real number, each term must be smaller than the whole sum; in particular, this implies that, for any non-negative integer $k$, we have $e^k > k^k / k!$.  Rearranging this slightly,
\begin{equation*}
1 < \frac{k! e^k}{k^k}
\end{equation*}
Multiplying this inequality by the previous inequality for the binomial coefficient yields
 \begin{equation*}
{n \choose k} < \frac{n^k}{k!} \cdot \frac{k! e^k}{k^k} = 
\left( \frac{ne}{k} \right)^k
\end{equation*}

To conclude the proof we show that
\begin{equation}
\label{eq:1}
\prod_{i=1}^{n-1} \left(1
+\frac{1}{i}\right)^i=\frac{n^n}{n!}\;\forall\;n \ge 2 \in
\mathbb{N}.
\end{equation}
\begin{eqnarray*}
\prod_{i=1}^{n-1} \left(1+\frac{1}{i}\right)^i&=&\prod_{i=1}^{n-1}
\frac{(i+1)^i}{i^i} \\
&=&\frac{\prod\limits_{i=2}^n i^{i-1}}{\left(\prod\limits_{i=1}^{n-1}
i^{i-1}\right)\cdot (n-1)!}
\end{eqnarray*}
Since each left-hand factor in (\ref{eq:1}) is $<e$, we have
$\frac{n^n}{n!} <e^{n-1}$.
Since $n-i<n\;\forall\;1 \le i \le k-1$, we immediately get
\begin{eqnarray*}
{n \choose k}&=&\frac{\prod\limits_{i=2}^{k-1}
\left(1-\frac{1}{i}\right)}{k!} <\frac{n^k}{k!}.
\end{eqnarray*}
And from
\begin{eqnarray*}
k \le n&\Leftrightarrow&(n-i)\cdot k \ge (k-i)\cdot n\;\forall \;1\le
i\le k-1
\end{eqnarray*}
we obtain
\begin{eqnarray*}
{n \choose k}&=&\frac{n}{k}\cdot \prod\limits_{i=1}^{k-1}
\frac{n-i}{k-i} \\ 
&\ge \left(\frac{n}{k}\right)^k.
\end{eqnarray*}
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\end{document}