# Convergence of arithmetic and geometric means of the n-th root of a sequence of certain binomial coefficients

#### Introduction

Blaise Pascal first introduced the triangle that would later come to hold his name [1], although in modern notations the so-called binomial coefficient denoted by $n \choose k$ may be more familiar to the reader. We shall prove a few interesting results regarding a sequence of the $n$-th root of means of the set of binomial coefficients
\label{binomial}
{n \choose 0}, {n \choose 1}, {n \choose 2}, {\cdots}, {n \choose n}.

In particular, if $A_n$ is the arithmetic mean of (\ref{binomial}) and $G_n$ is the geometric mean of (\ref{binomial}), we will show that the infinite sequences
\begin{equation*}
S_A = A_1, \sqrt{A_2}, \sqrt[3]{A_3}, \sqrt[4]{A_4}, \cdots, \quad S_G = G_1, \sqrt{G_2}, \sqrt[3]{G_3}, \sqrt[4]{G_4}, \cdots
\end{equation*}
converge to $2$ and $\sqrt{e}$ respectively.

#### Arithmetic mean

Let $P_n$ be the sum of the of sequence (\ref{binomial}), i.e.
\begin{equation*}
P_n = {n \choose 0} + {n \choose 1} + {n \choose 2} + {\cdots} + {n \choose n} = \sum\limits_{i=0}^{n}{n \choose i}.
\end{equation*}
Recall that
\begin{equation*}
{j \choose k} = {j-1 \choose k-1} + {j-1 \choose k}.
\end{equation*}
If we let $P_n = 0$ for $n < 0$ and let $P^m_n$ be the $m$th term of $P_n$, then $$P^m_n = P^{m-1}_{n-1} + P^{m}_{n-1},$$ which is to say that every term of $P_n$ is composed of terms of $P_{n-1}$. In fact, every term of $P_{n-1}$ is used twice by $P_n$; for example $P^0_{n} = P^0_{n-1}$, $P^1_{n} = P^0_{n-1} + P^1_{n-1}$, $P^2_{n} = P^1_{n-1} + P^2_{n-1}$, $P^3_{n} = P^2_{n-1} + P^3_{n-1}$. Since $P_n$ contains $n+1$ terms, all $n$ terms of $P_{n-1}$ will be used exactly twice, and therefore $$P_n = 2P_{n-1}.$$ Lastly since $P_0 = 1 = 2^0$ we arrive at $$\label{pn} P_n=2^n.$$ We wish to find the arithmetic mean of $P_n$, which contains $n+1$ terms, so $$A_n = \dfrac{P_n}{n+1}.$$ Equipped with (\ref{pn}) we can see that $S^n_A$, the $n$th term of the sequence $S_A$ we can see that $$S^n_A = \sqrt[n]{A_n}=\sqrt[n]{\dfrac{P_n}{n+1}} = \sqrt[n]{\dfrac{2^n}{n+1}}.$$ We wish to evaluate $$\lim\limits_{n \rightarrow \infty}{S_A} = \lim\limits_{n \rightarrow \infty}{\sqrt[n]{\dfrac{2^n}{n+1}}}.$$ We can reduce by first rewritting the limit as an exponential and applying the logarithm through: $$\lim\limits_{n \rightarrow \infty}{\sqrt[n]{\dfrac{2^n}{n+1}}} = \lim\limits_{n \rightarrow \infty}{e^{\ln\left(\left(\dfrac{2^n}{n+1}\right)^{\frac{1}{n}}\right)}} = \lim\limits_{n \rightarrow \infty}{e^{\dfrac{\ln\left(\frac{2^n}{n+1}\right)}{n}}}.$$ Pulling out the exponential function and expanding the logarithms, $$\lim\limits_{n \rightarrow \infty}{e^{\dfrac{\ln\left(\frac{2^n}{n+1}\right)}{n}}} = e^{\lim\limits_{n \rightarrow \infty}{\dfrac{n \ln(2) - \ln(n+1)}{n}}} = e^{\lim\limits_{n \rightarrow \infty}{\ln(2)}-\lim\limits_{n \rightarrow \infty}{\dfrac{\ln(n+1)}{n}}}$$ Since $n$ grows faster than $\ln(n+1)$, $\lim\limits_{n \rightarrow \infty}{\frac{\ln(n+1)}{n}}=0$, and $\ln(2)$ is a constant, $$e^{\lim\limits_{n \rightarrow \infty}{\ln(2)}-\lim\limits_{n \rightarrow \infty}{\dfrac{\ln(n+1)}{n}}} = e^{\ln(2)} = 2,$$ giving is the result that $$\lim\limits_{n \rightarrow \infty}{S_A} = 2.$$

#### Geometric mean

Let $Q_n$ be the product of the sequence (\ref{binomial}0):

Q_n = {n \choose 0}{n \choose 1}{n \choose 2}{\cdots}{n \choose n} = \prod\limits_{i=0}^{n}{n \choose i}.

Recall that

{n \choose k} = \dfrac{n!}{k!(n-k)!}.

So,

Q_n = \prod\limits_{i=0}^{n}{n \choose i} = \dfrac{n!}{0!(n-0)!}\dfrac{n!}{1!(n-1)!}\dfrac{n!}{2!(n-2)!}\cdots\dfrac{n!}{n!(n-n)!}.

We can pull out the constant factors in the denominator, leaving us with

Q_n = \dfrac{1}{0!1!2!\cdots n!}\left(\dfrac{n!}{(n-0)!}\dfrac{n!}{(n-1)!}\dfrac{n!}{(n-2)!}\cdots\dfrac{n!}{(n-n)!}\right).

Let us consider what happens when we divide out each factor inside the parenthesis. The first factor will be $1$, the second will be $n$, the third will be $n(n-1)$, the fourth will be $n(n-1)(n-2)$ and so forth. How many $n$s will this product have? Each factor (except the first) has an $n$ and $Q_n$ has $n+1$ of these factors, so the product will have $n$ number of $n$s, or $n^n$. Likewise for $(n-1)$ there will be $(n-1)$ factors with it, so the product will have $(n-1)^(n-1)$. Continuing this we arrive at
\label{qn_powers}
Q_n = \dfrac{n^n(n-1)^{n-1}(n-2)^{n-2}\cdots 2^2 1}{0!1!2!\cdots n!}.

The numerator of (\ref{qn_powers}) is the hyperfactorial [3], denoted as $H(n)$ . The denominator is the Barnes G-function [4], denoted as $G(n)$ for $n+2$. So, we can rewrite (\ref{qn_powers}) as
\label{qcompact}
Q_n = \dfrac{H(n)}{G(n+2)}.

The Barnes G-function is defined as
\label{gfunction}
G(n) = \dfrac{\Gamma(n)^{n-1}}{H(n-1)}.

Plugging (\ref{gfunction}) into (\ref{qcompact}), we now have

Q_n = \dfrac{H(n)}{G(n+2)} = \dfrac{H(n)}{\left[\dfrac{\Gamma(n+2)^{n+2-1}}{H(n+2-1)}\right]} = \dfrac{H(n)H(n+1)}{\Gamma(n+2)^{n+1}}.

Because $\Gamma(n)=(n-1)!$ when $n$ is a non-negative integer (which is always true for our $n$) the denominator can be simplified to

\dfrac{H(n)H(n+1)}{(n+1)!^{n+1}} = \dfrac{H(n)H(n)(n+1)^{n+1}}{n!^{n+1}(n+1)^{n+1}} = \dfrac{H(n)^2}{n!^{n+1}}.

We wish to find the geometric mean of (\ref{binomial}). The geometric mean of $n$ numbers is the product of the numbers to the $n$th root; $Q_n$ is the product of (\ref{binomial}), which has $n+1$ numbers, so the geometric mean $G_n$ of (\ref{binomial}) is

G_n=\sqrt[n+1]{Q_n} = \sqrt[n+1]{\dfrac{H(n)^2}{n!^{n+1}}} = \dfrac{H(n)^{\frac{2}{n+1}}}{n!}.

Ultimately we are interested in the convergence of the sequence $S_G = G_1,\sqrt{G_2},\sqrt[3]{G_3}$, so if $S^n_G$ is the $n$th term of this sequence, then

S^n_G = \sqrt[n]{G_n} = \sqrt[n]{\dfrac{H(n)^{\frac{2}{n+1}}}{n!}}.

Does $S_G$ converge? To answer this we will first consider a new sequence $S’$ where $S’_n$, the $n$th term of $S’$, is
\label{sprime}
S’_n = (S^n_G)^2.

In other words, the terms of $S’$ are the squares of the terms of $S_G$. Continuing,

S’_n = (S^n_G)^2 = \left(\dfrac{H(n)^{\frac{2}{n+1}}}{n!}\right)^{\frac{2}{n}} = \dfrac{H(n)^{\frac{4}{n^2+n}}}{n!^{\frac{2}{n}}}

Now is a convenient time to mention [5]:
\label{e}
\lim\limits_{n \rightarrow \infty}{\dfrac{n}{\sqrt[n]{n!}}} = e.

If we take the limit of the difference of (\ref{e}) and $S’$, we find

\lim\limits_{n \rightarrow \infty}{\left(\dfrac{n}{\sqrt[n]{n!}} – \dfrac{H(n)^{\frac{4}{n^2+n}}}{n!^{\frac{2}{n}}}\right)} =
\lim\limits_{n \rightarrow \infty}{\left(\dfrac{n\sqrt[n]{n!} – H(n)^{\frac{4}{n^2+n}}}{n!^{\frac{2}{n}}}\right)} = 0.

Since the difference of the two limits is zero, we can conclude
\label{squarede}
\lim\limits_{n \rightarrow \infty}{\left(\dfrac{n}{\sqrt[n]{n!}} – \dfrac{H(n)^{\frac{4}{n^2+n}}}{n!^{\frac{2}{n}}}\right)} = 0 \Rightarrow
\lim\limits_{n \rightarrow \infty}{\dfrac{n}{\sqrt[n]{n!}}} = \lim\limits_{n \rightarrow \infty}\dfrac{H(n)^{\frac{4}{n^2+n}}}{n!^{\frac{2}{n}}} \Rightarrow \lim\limits_{n \rightarrow \infty}\dfrac{H(n)^{\frac{4}{n^2+n}}}{n!^{\frac{2}{n}}} = e.

$S_G$ is bounded above by $S’$, so it converges and by (\ref{squarede}) and (\ref{sprime})

\lim\limits_{n \rightarrow \infty}{S’} = \lim\limits_{n \rightarrow \infty}{(S_G)^2} \Rightarrow
\lim\limits_{n \rightarrow \infty}{S’} = \left(\lim\limits_{n \rightarrow \infty}{S_G}\right)^2 \Rightarrow e = \left(\lim\limits_{n \rightarrow \infty}{S_G}\right)^2.

Finally,

\lim\limits_{n \rightarrow \infty}{S_G} = \sqrt{e}.

[1] Katz, Victor J. A History of Mathematics: An Introduction. Boston: Addison-Wesley, 2009.
[2] Young, Robert M. Excursions in Calculus: An Interplay of the Continuous and the Discrete. Washington, D.C.: The Mathematical Association of America, 1992.
[3] Azarian, M K. 2007. On the hyperfactorial function, hypertriangular function, and discriminants of certain polynomials. International Journal of Pure and Applied Mathematics 36, (2): 249-255
[4] Barnes, E. W. 1900. The theory of the G-function. Quarterly Journal of Pure and Applied Mathematics 31, 264-314.
[5] Weisstein, Eric W. Stirling’s Approximation.” From MathWorld–A Wolfram Web Resource. http://mathworld.wolfram.com/StirlingsApproximation.html

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