Starting Later: Geometric Bounds for Stirling’s Formula

Mohan Rajendran

Stirling's formula describes the extraordinary growth of a factorial with a remarkably compact expression:

n!\sim\sqrt{2\pi n}\left(\frac ne\right)^n

This also means that

R_n:=\frac{n!}{(n/e)^n\sqrt n}\longrightarrow\sqrt{2\pi}

In “Stirling's Formula: A Better Approximation”, Bikash Chakraborty returns to an integral comparison from Walter Rudin's Principles of Mathematical Analysis. Rudin's exercise (Chapter 8, Problem 20) gives $e^{7/8}<R_n<e$ for integers $n>2$ . Chakraborty keeps the same basic idea, modifies the geometry, and sharpens both bounds to

e^{7/8}<\left(\frac{2e}{3}\right)^{3/2}<R_n<\frac{e^2}{2\sqrt2}<e,\qquad n>2

In decimals,

2.39887529397<2.43952253514<R_n<2.61242583706<2.71828182846

The proof is highly pedagogical because it turns a factorial into a comparison of visible areas under $y=\log x$ .

What caught my attention was a small geometric choice on which the argument depends: the point at which the area comparison begins. Chakraborty begins at $2$ . If the construction begins later, the initial factors can be retained exactly while only the remaining tail is estimated.

The geometric core idea of Chakraborty’s argument

For completeness, let us try to understand Chakraborty's argument. Taking logarithms turns the product defining $n!$ into the sum $\log(n!)=\sum_{j=1}^n\log j$ . If $g(x)=\log x$ , then $g''(x)=-1/x^2<0$ , so the graph is strictly concave. Every chord lies below the curve, while every tangent line lies above it. These facts give two complementary area comparisons.

On $[m,m+1]$ , the chord joining $(m,\log m)$ and $(m+1,\log(m+1))$ lies below the curve. For the other direction, the tangent at the integer $m$ lies above the curve on the centred interval $[m-\tfrac12,m+\tfrac12]$ . Thus the tangent points are integers and the endpoints of their intervals are half-integers. A final half-width rectangle completes the cover up to $n$ .

Interactive 1 · Chakraborty's construction

The areas in the proof

Endpoint

n

6

Visible areasArea under log xChord trapezoidsTangent cover

Chords belowdomain [2, n]

$C_n<\int_2^n\log x\,dx$

5.33680 < 5.36426gap 0.02746

Tangents abovedomain [3/2, n]

$\int_{3/2}^n\log x\,dx<T_n$

5.64236 < 5.68337gap 0.04101

At $n=6$ , the chord total undercounts by 0.02746, while the tangent cover overcounts by 0.04101. The two panels use different starting points, exactly as in the proof.

Chord trapezoids and the upper bound

For an integer $m$ , the chord has equation

\ell_m(x)=(m+1-x)\log m+(x-m)\log(m+1),\qquad m\le x\le m+1

The interval has width one, and the two vertical sides have heights $\log m$ and $\log(m+1)$ . Its area is therefore the trapezoid area

\int_m^{m+1}\ell_m(x)\,dx=\frac{\log m+\log(m+1)}{2}

For Chakraborty's starting point $k=2$ , add these trapezoids for $m=2,\ldots,n-1$ . Each intermediate logarithm occurs twice with coefficient one half, while the two endpoint logarithms occur once. Hence

\begin{aligned}C_n&=\sum_{m=2}^{n-1}\frac{\log m+\log(m+1)}2\\&=\frac12\log2+\sum_{m=3}^{n-1}\log m+\frac12\log n\\&=\log(n!)-\frac12\log(2n)<\int_2^n\log x\,dx.\end{aligned}

Now $\int_2^n\log x\,dx=n\log n-n-2\log2+2$ . Substituting this and collecting the terms in $\log R_n$ gives

\log R_n<2-\frac32\log2=\log\!\left(\frac{e^2}{2\sqrt2}\right),\qquad R_n<\frac{e^2}{2\sqrt2}.

Tangent cells and the lower bound

The tangent to $\log x$ at the integer $m$ is

\tau_m(x)=\log m+\frac{x-m}{m}.

Chakraborty uses it on the unit cell centred at $m$ . Put $u=x-m$ . The constant part contributes $\log m$ , while the linear part has equal positive and negative areas and cancels:

\begin{aligned}\int_{m-1/2}^{m+1/2}\tau_m(x)\,dx&=\int_{-1/2}^{1/2}\left(\log m+\frac{u}{m}\right)du\\&=\log m.\end{aligned}

This is the tangent-cell formula. The cells for $m=2,\ldots,n-1$ cover the interval from $3/2$ to $n-1/2$ . To reach $n$ , add a final rectangle of width one half and height $\log n$ . Their combined area is

\begin{aligned}T_n&=\sum_{m=2}^{n-1}\log m+\frac12\log n\\&=\log(n!)-\frac12\log n.\end{aligned}

Because every tangent cell and the final rectangle lie above the logarithmic curve,

\int_{3/2}^{n}\log x\,dx<T_n=\log(n!)-\frac12\log n.

Evaluating the integral and rearranging gives

\log R_n>\frac32-\frac32\log\frac32=\log\!\left(\frac{2e}{3}\right)^{3/2},\qquad \left(\frac{2e}{3}\right)^{3/2}<R_n.

The two constants therefore record the same concave curve in two ways: the chord construction begins at $2$ , while the first centred tangent cell begins at $3/2$ .

Move the starting point

Fix integers $n\ge k\ge2$ . Keep the initial product $1\cdot2\cdots(k-1)=(k-1)!$ untouched, and apply the geometric comparison only from $k$ onwards.

Interactive 2 · assemble the proof

Move the starting line

Start

k

3End

n

8

Chords belowan under-area makes the upper bound

Gap 0.01727

Tangents abovean over-area makes the lower bound

Gap 0.02693

Chord construction $R_{8}<R_{3}=2.576976$

Tangent construction $2.465563=L_{3}<R_{8}$

Only the starting line moves. The factors before $k$ remain exact; the same local comparison is applied to the tail.

For $n>k$ , adding the chord trapezoids from $k$ to $n$ gives

\frac12\log k+\sum_{j=k+1}^{n-1}\log j+\frac12\log n<\int_k^n\log x\,dx

Evaluating the integral and collecting logarithms yields $R_n<R_k$ . When $n=k$ , equality is immediate. Hence $R_n\le R_k$ , and the upper constant at starting point $k$ is $U_k=R_k=\dfrac{k!e^k}{k^k\sqrt k}$ .

Add the centred tangent cells for $j=k,\ldots,n-1$ and finish with a half-width rectangle of height $\log n$ . Their combined area lies strictly above the curve, so

\int_{k-1/2}^{n}\log x\,dx<\sum_{j=k}^{n-1}\log j+\frac12\log n

Rearrangement gives $L_k=(k-1)!\left(\dfrac{e}{k-\tfrac12}\right)^{k-1/2}$ .

A family of bounds

The two constructions combine into the following theorem.

At $k=2$ , this recovers Chakraborty's pair of constants. Moving the starting point one step, to $k=3$ , gives for every integer $n>2$

2\left(\frac{2e}{5}\right)^{5/2}<R_n\le\frac{2e^3}{9\sqrt3}

or $2.465563424<R_n\le2.576975589$ . This interval is about $35.6\%$ narrower than Chakraborty's interval over the same integer range. The non-strict upper sign matters: equality occurs at $n=3$ .

Starting point	Uniform tail	Lower endpoint	Upper endpoint	Width
$k=2$	$n\ge2$	2.439522535	2.612425837	0.172903302
$k=3$	$n\ge3$	2.465563424	2.576975589	0.111412165
$k=5$	$n\ge5$	2.483591120	2.548699488	0.065108369
$k=10$	$n\ge10$	2.495665416	2.527597120	0.031931705
$k=20$	$n\ge20$	2.501278769	2.517093475	0.015814706

When the integer becomes real

The Gamma function extends the factorial beyond the integers. Since $\Gamma(n+1)=n!$ , define, for real $x>2$ ,

R(x):=\frac{\Gamma(x+1)}{(x/e)^x\sqrt x}

At an integer $n$ , this is exactly $R_n$ . The lower endpoints of the starting-point construction also have the continuous companion

L(x):=\Gamma(x)\left(\frac{e}{x-\tfrac12}\right)^{x-1/2}

with $L(k)=L_k$ and $R(k)=U_k$ at every integer $k\ge2$ . The comparison is transparent because the Gamma factors cancel:

\frac{R(x)}{L(x)}=\sqrt e\left(1-\frac1{2x}\right)^{x-1/2}>1

The inequality still has the same geometric meaning. In fact,

\begin{aligned}\log\frac{R(x)}{L(x)}&=\frac12\log x-\int_{x-1/2}^{x}\log t\,dt\\&=\int_{x-1/2}^{x}\log\frac{x}{t}\,dt>0\end{aligned}

This is the excess of the final half-width rectangle of height $\log x$ above the logarithmic curve. Consequently $L(x)<R(x)$ for every real $x>2$ .

Interactive 3 · the continuous picture

Between the integers

Probe

x

6.00

R(x)L(x)√2π

Lower $L(x)$ 2.487746388

Upper $R(x)$ 2.541653013

Gap0.053906626

At $x=6.00$ , the ratio $R(x)/L(x)$ is 1.0216689. The shaded interval narrows towards $\sqrt{2\pi}$ .

The quotient $R(x)/L(x)$ tends to $1$ . The Gamma-function form of Stirling's formula gives $R(x)\to\sqrt{2\pi}$ , and so $L(x)$ approaches the same limit. Here $L(x)$ is a continuous companion to the geometric lower endpoints, not an independent computational estimate for Gamma: away from the integers, the original sum of logarithms no longer telescopes into a factorial.

Sources and further reading

Bikash Chakraborty, “Stirling's Formula: A Better Approximation”, The College Mathematics Journal, published online 31 August 2026, DOI 10.1080/07468342.2026.2721230.
Walter Rudin, Principles of Mathematical Analysis, 3rd ed., Chapter 8, Problem 20.
NIST Digital Library of Mathematical Functions, §5.11, Asymptotic Expansions, for the Gamma-function form of Stirling's expansion.

The approximation itself did not change. The sharper bounds appeared when the same geometry was asked to begin somewhere else.

Article information

Keywords

Stirling’s formula · asymptotic analysis · inequalities · Stirling approximation

Mathematics Subject Classification (2020)

Primary41A60Asymptotic approximations and expansions
Secondary26D15Inequalities for sums, series and integrals

Cite this article

Rajendran, Mohan. “Starting Later: Geometric Bounds for Stirling’s Formula.” Math Nomad, 7 September 2026, https://mathnomad.in/articles/moving-the-starting-line-in-stirlings-formula/

BibLaTeX

@online{rajendranStirlingBounds,
  author       = {Rajendran, Mohan},
  title        = {Starting Later: Geometric Bounds for Stirling’s Formula},
  date         = {2026-09-07},
  organization = {Math Nomad},
  url          = {https://mathnomad.in/articles/moving-the-starting-line-in-stirlings-formula/},
  langid       = {british}
}