Asymptotic equivalence ($a_n\sim b_n⟺a_n/b_n\to 1$)

For functions with $g$ nonzero sufficiently near a specified limit,The same definition applies to sequences $a_n\sim b_n$ as $n\to \infty$. Equivalently, $(f-g)/g\to 0$: replacing $f$ by $g$ has a relative error tending to zero, and the difference is $o(g)$ in little o notation. The limit and reference scale are essential parts of the statement.

For example, $n^2+n\sim n^2$ because $(n^2+n)/n^2=1+1/n\to 1$. Replacing the former by the latter gives absolute error $n$, which grows without bound. Thus asymptotic equivalence controls a ratio, rather than requiring the absolute error to vanish. This distinction is crucial when approximating a rapidly growing factorial with Stirling's formula.

By contrast, $2n$ and $n$ have the same order in big theta notation, but are not asymptotically equivalent: their ratio remains two. For positive functions, asymptotic equivalence gives both-sided constant bounds eventually, whereas constant bounds alone do not recover the leading coefficient.

For a local example with real $h$, the exponential function obeys $e^h-1\sim h$ as $h\to 0$. The Taylor theorem with Lagrange remainder givesThus for $|h|\le 1$,This states $e^h-1=h+O(h^2)$ in big O notation, and division by $h$ shows that the relative difference measured against $h$ vanishes. The diagram uses positive $h$; the bound is valid on both sides of zero. At $h=0.1$, the absolute error is approximately $0.00517092$ and the relative difference against $h$ is approximately $0.0517092$.

For positive bases $a,b$, comparison of the exponential functions $a^n$ and $b^n$ reduces to a geometric progression:If $a<b$, this ratio tends to zero, so $a^n=o(b^n)$ in little o notation. If $a=b$, the ratio is one; if $a>b$, it grows without bound.

Consequently, provided at least one coefficient is nonzero, a finite sum with distinct positive bases is asymptotically equivalent to its largest-base term whose coefficient is nonzero. For example,The relative difference against $5\cdot 3^n$ first drops below one percent at the integer $n=11$. Dividing by the proposed dominant term provides an explicit error estimate, not just a comparison of two rapidly growing numbers. If several terms have the same largest base, their coefficients must first be combined; cancellation can remove that base entirely.

Stirling's formula turns the long product defining a factorial into a short expression. For positive integers $n$, putThen $n!\sim S_n$ as $n\to \infty$. This is asymptotic equivalence: $S_n/n!\to 1$, so the relative error tends to zero. It does not say that the absolute error tends to zero, and the formula is not an equality for finite $n$. The Gamma function extends this approximation beyond integer inputs. ^[ref]

For $n=10$,The leading estimate is about $0.830\%$ low. Multiplying by the first correction factor giveswhich is about $0.003176\%$ low. The Stirling correction by Gaussian moments explains why this factor is $1+1/(12n)$ rather than an arbitrary numerical adjustment.

The usefulness is in ratios as well as in enormous numbers. Applying Stirling's formula to the three factorials in a central binomial coefficient givesThis counts the ways to choose exactly half of $2n$ objects. Equivalently, for $2n$ independent fair coin tosses, the binomial distribution assigns probability $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)/4^n\sim 1/\sqrt{\pi n}$ to exactly $n$ heads. An exponentially large count becomes a simple probability scale after cancellation.

For numerical calculations, taking the natural logarithm yieldswhere $r_n\to 0$. Computing a logarithm of a large factorial avoids representing the factorial itself. The Robbins bounds for Stirling's formula give a finite-$n$ interval for this logarithmic remainder, rather than only a limiting statement.

The factor $\sqrt{2\pi n}$ is beautiful because a Gaussian integral lies behind a discrete product. The Gaussian peak behind Stirling's formula displays this connection through the Gamma integral and Laplace's method. A rigorous proof must also control the tails of the integral; the local expansion explains the geometry but is not by itself a complete proof.

James Stirling published Methodus differentialis in 1730, a treatise on summation and interpolation of infinite series. This setting connects the formula with a practical calculation problem: taking logarithms turns a long product into a sum, whose growth can be approximated. ^[ref]

For integer $N\ge 2$, write $\ln (N!)={\sum }_{k=2}^N\ln k$. Monotonicity gives ${\int }_{k-1}^k\ln xdx\le \ln k\le {\int }_k^{k+1}\ln xdx$. Summing and evaluating the endpoint integrals proves the displayed bounds. They also hold with equality when $N=1$. The lower integral starts at one and the upper at two; confusing those endpoints loses the constant in the upper bound. The bounds imply $\ln (N!)=N\ln N-N+O(\ln N)$, while Stirling's formula determines the logarithmic and constant corrections.

The positive sequence $a_n=\sqrt{n}(n/e{)}^n/n!$ has logarithmic incrementsThe rational lower bound for the natural logarithm proves $u_n>0$. For $n>1$, the Taylor series givesSince $0<(k-1)/[2k(k+1)]<1/2$, a geometric series bounds $u_n<1/[2n(n-1)]\le n^{-2}$ for $n\ge 2$. The comparison test for series therefore proves $\sum u_n<\infty$. The finite telescoping sum $\ln a_{n+1}=-1+{\sum }_{k=1}^n{u}_k$ shows that $a_n$ increases to a finite positive limit. This argument establishes the existence of the constant in Stirling's formula; its value $1/\sqrt{2\pi }$ requires an independent identification.

For every positive integer $n$, Herbert Robbins's bounds giveEquivalently, the remainder in $\log (n!)=\log S_n+r_n$ satisfies $1/(12n+1)<r_n<1/(12n)$. This both proves that $S_n$ underestimates $n!$ and squeezes $n!/S_n$ to one, establishing Stirling's formula. Robbins published the bounds in “A Remark on Stirling's Formula” in 1955. ^[ref]

For $n=10$, the relative error of $S_{10}$ lies strictly between $1-e^{-1/121}$ and $1-e^{-1/120}$, approximately $0.823\%$ and $0.830\%$. A bound specifies the accuracy of an approximation before using it in another calculation.

Repeated integration by parts gives the Gamma integralThe logarithm of its integrand is $n\log t-t$, whose unique maximum is at $t=n$. Write $t=n+\sqrt{n}s$. For fixed $s$ and large $n$, a Taylor expansion givesThe Big O notation here is for a fixed $s$, or uniformly for $s$ in a fixed bounded interval. It does not claim a uniform approximation over the entire growing integration domain. Consequently the peak, divided by its height, approaches $e^{-s^2/2}$.

The exact substitution givesReplacing the integral by the Gaussian integral ${\int }_{\mathbb{R}}{e}^{-s^2/2}ds=\sqrt{2\pi }$ produces $S_n$. Thus the estimate is peak height $n^n{e}^{-n}$ times width $\sqrt{n}$ times the area of a standard Gaussian peak. This is Laplace's method in a concrete form; controlling the integral's tails justifies the replacement. The next terms, including their products after exponentiation, produce the Stirling correction by Gaussian moments.

In Laplace method for the Gamma function, scaling $t=x(1+s/\sqrt{x})$ gives a standard Gaussian integral. The first even correction polynomial is $s^2-7s^4/12+s^6/18$. Its expectation under a unit-variance centered Gaussian is $1-7\cdot 3/12+15/18=1/12$. The amplitude expansion contributes to this coefficient as well as the phase expansion; expanding only the latter gives the wrong Stirling formula correction.