Show that Gamma(x)=frac1x product from n=1 to infinity
Show that $\displaystyle\Gamma(x)=\frac1x\prod_{n=1}^{\infty}\left(1+\frac1n\right)^x\left(1+\frac xn\right)^{-1}$ converges absolutely for every $x$ other than $0$ and the negative integers.
Explanation of Method
The key is that the $O(1/n)$ leading terms of $x\ln(1+1/n)$ and $\ln(1+x/n)$ cancel exactly when subtracted, leaving a general term of size $O(1/n^2)$ — comfortably summable regardless of $x$. Excluding $x=0$ (the prefactor $1/x$) and negative integers (where some factor $1+x/n$ vanishes) are the only obstructions.
- Expand $\ln a_n=x\ln(1+1/n)-\ln(1+x/n)$ to second order in $1/n$.
- Show the leading $O(1/n)$ terms cancel, leaving $O(1/n^2)$.
- Conclude absolute convergence via comparison with $\sum1/n^2$, and identify the excluded points.
Step 1 · Second-order expansion
With $a_n=\dfrac{(1+1/n)^x}{1+x/n}$: $$ \ln a_n = x\ln\left(1+\frac1n\right)-\ln\left(1+\frac xn\right) = x\left(\frac1n-\frac1{2n^2}+O\!\left(\frac1{n^3}\right)\right) - \left(\frac xn-\frac{x^2}{2n^2}+O\!\left(\frac1{n^3}\right)\right). $$
Step 2 · The leading terms cancel
The $O(1/n)$ terms are both $x/n$ and cancel exactly: $$ \ln a_n = \left(\frac{x^2}{2n^2}-\frac x{2n^2}\right)+O\!\left(\frac1{n^3}\right) = \frac{x(x-1)}{2n^2}+O\!\left(\frac1{n^3}\right). $$
Step 3 · Absolute convergence
Since $|\ln a_n|\sim\dfrac{|x(x-1)|}{2n^2}$ and $\sum1/n^2=\pi^2/6<\infty$, comparison gives $\sum|\ln a_n|<\infty$ for every fixed $x$, so $$ \boxed{\prod_{n=1}^{\infty}|a_n|\ \text{converges for every }x, \text{ hence the defining product for }\Gamma(x)\text{ converges absolutely.}} $$ The prefactor $1/x$ excludes $x=0$; at $x=-k$ ($k\in\mathbb N^+$), the factor at $n=k$ is $1+x/n=1-k/k=0$, making the product (and hence $\Gamma$) undefined there. For every other $x$, all factors are nonzero and the product converges absolutely.
Summary Table
| Property | Value |
|---|---|
| General term | $\ln a_n=\dfrac{x(x-1)}{2n^2}+O(1/n^3)$ |
| Convergence | absolute, for all $x\in\mathbb R\setminus\{0,-1,-2,\ldots\}$ |
Additional Notes
- This is the Weierstrass-style product definition of the Gamma function, equivalent to (but distinct in form from) the Euler limit formula $\Gamma(x)=\lim_{n\to\infty}\dfrac{n!\,n^x}{x(x+1)\cdots(x+n)}$ derived in Problem 388 — the two are algebraically the same object viewed as an infinite product versus a limit of partial products.
- The exact cancellation of the $O(1/n)$ terms here is structurally identical to Problem 376's cancellation of the $O(x^2)$ term in $\cosh x\cos x$: in both cases, subtracting two Taylor expansions designed to nearly match removes the leading-order behavior, leaving a much faster-converging remainder.
- The excluded points $x=0,-1,-2,\ldots$ are exactly the poles of $\Gamma(x)$, consistent with the classical fact that $\Gamma$ has simple poles at the non-positive integers and is otherwise analytic on all of $\mathbb C$.