Blaise Pascal first introduced the triangle that would later come to hold his name [1<\/a>], 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 <\/p>\n Let $P_n$ be the sum of the of sequence (\\ref{binomial}), i.e. Let $Q_n$ be the product of the sequence (\\ref{binomial}0): [1] Katz, Victor J. A History of Mathematics: An Introduction. Boston: Addison-Wesley, 2009. <\/a> 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 […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":[],"categories":[7],"tags":[],"_links":{"self":[{"href":"http:\/\/jeffq.com\/blog\/wp-json\/wp\/v2\/posts\/1380"}],"collection":[{"href":"http:\/\/jeffq.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"http:\/\/jeffq.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"http:\/\/jeffq.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"http:\/\/jeffq.com\/blog\/wp-json\/wp\/v2\/comments?post=1380"}],"version-history":[{"count":10,"href":"http:\/\/jeffq.com\/blog\/wp-json\/wp\/v2\/posts\/1380\/revisions"}],"predecessor-version":[{"id":1397,"href":"http:\/\/jeffq.com\/blog\/wp-json\/wp\/v2\/posts\/1380\/revisions\/1397"}],"wp:attachment":[{"href":"http:\/\/jeffq.com\/blog\/wp-json\/wp\/v2\/media?parent=1380"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/jeffq.com\/blog\/wp-json\/wp\/v2\/categories?post=1380"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/jeffq.com\/blog\/wp-json\/wp\/v2\/tags?post=1380"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\\begin{equation}\\label{binomial}
\n{n \\choose 0}, {n \\choose 1}, {n \\choose 2}, {\\cdots}, {n \\choose n}.
\n\\end{equation}
\nIn 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
\n\\begin{equation*}
\nS_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
\n\\end{equation*}
\nconverge to $2$ and $\\sqrt{e}$ respectively.<\/p>\nArithmetic mean<\/h4>\n
\n\\begin{equation*}
\nP_n = {n \\choose 0} + {n \\choose 1} + {n \\choose 2} + {\\cdots} + {n \\choose n} = \\sum\\limits_{i=0}^{n}{n \\choose i}.
\n\\end{equation*}
\nRecall that
\n\\begin{equation*}
\n{j \\choose k} = {j-1 \\choose k-1} + {j-1 \\choose k}.
\n\\end{equation*}
\nIf we let $P_n = 0$ for $n < 0$ and let $P^m_n$ be the $m$th term of $P_n$, then\n\\begin{equation}\nP^m_n = P^{m-1}_{n-1} + P^{m}_{n-1},\n\\end{equation}\nwhich 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 \n\\begin{equation}\nP_n = 2P_{n-1}.\n\\end{equation}\nLastly since $P_0 = 1 = 2^0$ we arrive at \n\\begin{equation} \\label{pn}\nP_n=2^n.\n\\end{equation}\nWe wish to find the arithmetic mean of $P_n$, which contains $n+1$ terms, so\n\\begin{equation}\nA_n = \\dfrac{P_n}{n+1}.\n\\end{equation}\nEquipped with (\\ref{pn}) we can see that $S^n_A$, the $n$th term of the sequence $S_A$ we can see that\n\\begin{equation}\nS^n_A = \\sqrt[n]{A_n}=\\sqrt[n]{\\dfrac{P_n}{n+1}} = \\sqrt[n]{\\dfrac{2^n}{n+1}}.\n\\end{equation}\nWe wish to evaluate\n\\begin{equation}\n\\lim\\limits_{n \\rightarrow \\infty}{S_A} = \\lim\\limits_{n \\rightarrow \\infty}{\\sqrt[n]{\\dfrac{2^n}{n+1}}}.\n\\end{equation}\nWe can reduce by first rewritting the limit as an exponential and applying the logarithm through:\n\\begin{equation}\n\\lim\\limits_{n \\rightarrow \\infty}{\\sqrt[n]{\\dfrac{2^n}{n+1}}} = \n\\lim\\limits_{n \\rightarrow \\infty}{e^{\\ln\\left(\\left(\\dfrac{2^n}{n+1}\\right)^{\\frac{1}{n}}\\right)}} = \n\\lim\\limits_{n \\rightarrow \\infty}{e^{\\dfrac{\\ln\\left(\\frac{2^n}{n+1}\\right)}{n}}}.\n\\end{equation}\nPulling out the exponential function and expanding the logarithms,\n\\begin{equation}\n\\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}}} = \ne^{\\lim\\limits_{n \\rightarrow \\infty}{\\ln(2)}-\\lim\\limits_{n \\rightarrow \\infty}{\\dfrac{\\ln(n+1)}{n}}}\n\\end{equation}\nSince $n$ grows faster than $\\ln(n+1)$, $\\lim\\limits_{n \\rightarrow \\infty}{\\frac{\\ln(n+1)}{n}}=0$, and $\\ln(2)$ is a constant,\n\\begin{equation}\ne^{\\lim\\limits_{n \\rightarrow \\infty}{\\ln(2)}-\\lim\\limits_{n \\rightarrow \\infty}{\\dfrac{\\ln(n+1)}{n}}} = e^{\\ln(2)} = 2,\n\\end{equation}\ngiving is the result that\n\\begin{equation}\n\\lim\\limits_{n \\rightarrow \\infty}{S_A} = 2.\n\\end{equation}\n\n\n\nGeometric mean<\/h4>\n
\n\\begin{equation}
\nQ_n = {n \\choose 0}{n \\choose 1}{n \\choose 2}{\\cdots}{n \\choose n} = \\prod\\limits_{i=0}^{n}{n \\choose i}.
\n\\end{equation}
\nRecall that
\n\\begin{equation}
\n{n \\choose k} = \\dfrac{n!}{k!(n-k)!}.
\n\\end{equation}
\nSo,
\n\\begin{equation}
\nQ_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)!}.
\n\\end{equation}
\nWe can pull out the constant factors in the denominator, leaving us with
\n\\begin{equation}
\nQ_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).
\n\\end{equation}
\nLet 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
\n\\begin{equation}\\label{qn_powers}
\nQ_n = \\dfrac{n^n(n-1)^{n-1}(n-2)^{n-2}\\cdots 2^2 1}{0!1!2!\\cdots n!}.
\n\\end{equation}
\nThe numerator of (\\ref{qn_powers}) is the hyperfactorial<\/em> [3<\/a>], denoted as $H(n)$ . The denominator is the Barnes G-function<\/em> [4<\/a>], denoted as $G(n)$ for $n+2$. So, we can rewrite (\\ref{qn_powers}) as
\n\\begin{equation}\\label{qcompact}
\nQ_n = \\dfrac{H(n)}{G(n+2)}.
\n\\end{equation}
\nThe Barnes G-function is defined as
\n\\begin{equation}\\label{gfunction}
\nG(n) = \\dfrac{\\Gamma(n)^{n-1}}{H(n-1)}.
\n\\end{equation}
\nPlugging (\\ref{gfunction}) into (\\ref{qcompact}), we now have
\n\\begin{equation}
\nQ_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}}.
\n\\end{equation}
\nBecause $\\Gamma(n)=(n-1)!$ when $n$ is a non-negative integer (which is always true for our $n$) the denominator can be simplified to
\n\\begin{equation}
\n\\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}}.
\n\\end{equation}
\nWe 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
\n\\begin{equation}
\nG_n=\\sqrt[n+1]{Q_n} = \\sqrt[n+1]{\\dfrac{H(n)^2}{n!^{n+1}}} = \\dfrac{H(n)^{\\frac{2}{n+1}}}{n!}.
\n\\end{equation}
\nUltimately 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
\n\\begin{equation}
\nS^n_G = \\sqrt[n]{G_n} = \\sqrt[n]{\\dfrac{H(n)^{\\frac{2}{n+1}}}{n!}}.
\n\\end{equation}
\nDoes $S_G$ converge? To answer this we will first consider a new sequence $S’$ where $S’_n$, the $n$th term of $S’$, is
\n\\begin{equation}\\label{sprime}
\nS’_n = (S^n_G)^2.
\n\\end{equation}
\nIn other words, the terms of $S’$ are the squares of the terms of $S_G$. Continuing,
\n\\begin{equation}
\nS’_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}}}
\n\\end{equation}
\nNow is a convenient time to mention [5<\/a>]:
\n\\begin{equation}\\label{e}
\n\\lim\\limits_{n \\rightarrow \\infty}{\\dfrac{n}{\\sqrt[n]{n!}}} = e.
\n\\end{equation}
\nIf we take the limit of the difference of (\\ref{e}) and $S’$, we find
\n\\begin{equation}
\n\\lim\\limits_{n \\rightarrow \\infty}{\\left(\\dfrac{n}{\\sqrt[n]{n!}} – \\dfrac{H(n)^{\\frac{4}{n^2+n}}}{n!^{\\frac{2}{n}}}\\right)} =
\n\\lim\\limits_{n \\rightarrow \\infty}{\\left(\\dfrac{n\\sqrt[n]{n!} – H(n)^{\\frac{4}{n^2+n}}}{n!^{\\frac{2}{n}}}\\right)} = 0.
\n\\end{equation}
\nSince the difference of the two limits is zero, we can conclude
\n\\begin{equation} \\label{squarede}
\n\\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
\n\\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.
\n\\end{equation}
\n$S_G$ is bounded above by $S’$, so it converges and by (\\ref{squarede}) and (\\ref{sprime})
\n\\begin{equation}
\n\\lim\\limits_{n \\rightarrow \\infty}{S’} = \\lim\\limits_{n \\rightarrow \\infty}{(S_G)^2} \\Rightarrow
\n\\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.
\n\\end{equation}
\nFinally,
\n\\begin{equation}
\n\\lim\\limits_{n \\rightarrow \\infty}{S_G} = \\sqrt{e}.
\n\\end{equation}<\/p>\n
\n[2] Young, Robert M. Excursions in Calculus: An Interplay of the Continuous and the Discrete. Washington, D.C.: The Mathematical Association of America, 1992. <\/a>
\n[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 <\/a>
\n[4] Barnes, E. W. 1900. The theory of the G-function. Quarterly Journal of Pure and Applied Mathematics 31, 264-314. <\/a>
\n[5] Weisstein, Eric W. “Stirling’s Approximation.” From MathWorld–A Wolfram Web Resource. http:\/\/mathworld.wolfram.com\/StirlingsApproximation.html<\/a><\/a><\/p>\n","protected":false},"excerpt":{"rendered":"