Prove that Gamma(x)=lim as n to infinity (n!
Prove that $\displaystyle\Gamma(x)=\lim_{n\to\infty}\frac{n!\,n^x}{x(x+1)(x+2)\cdots(x+n)}$. Then show $\Gamma(x+1)=x\Gamma(x)$, and verify $\Gamma(n+1)=n!$.
Explanation of Method
Starting from Problem 387's product definition, the partial product telescopes exactly into a ratio of factorials, yielding Euler's limit formula directly. The functional equation then follows by inserting $x+1$ into the limit formula, multiplying through by $x$, and observing the extra factor $n/(x+n+1)\to1$.
- Telescope the partial product from Problem 387 into $P_N=(N+1)^x\,N!/[x(x+1)\cdots(x+N)]$.
- Take the limit to recover Euler's formula.
- Derive $\Gamma(x+1)=x\Gamma(x)$ directly from the limit formula, then compute $\Gamma(1)=1$ and conclude $\Gamma(n+1)=n!$.
Step 1 · Telescoping the partial product
From Problem 387, $$ P_N=\frac1x\prod_{n=1}^N\frac{(1+1/n)^x}{1+x/n} = \frac1x\prod_{n=1}^N\frac{(n+1)^x\,n}{n^x(n+x)}. $$ The numerator telescopes: $\prod_{n=1}^N(n+1)^x=\bigl((N+1)!\bigr)^x$ and $\prod_{n=1}^Nn=N!$. The denominator: $\prod_{n=1}^Nn^x=(N!)^x$ and $\prod_{n=1}^N(n+x)=(x+1)(x+2)\cdots(x+N)$. Since $(N+1)!^x/(N!)^x=(N+1)^x$: $$ P_N = \frac{(N+1)^x\,N!}{x(x+1)(x+2)\cdots(x+N)}. $$ (Verified directly at $x=2,N=2$: $P_2=3^2\cdot2!/(2\cdot3\cdot4)=18/24=0.75$, matching the product computed term-by-term.)
Step 2 · Euler's limit formula
Since $(N+1)^x\sim N^x$ as $N\to\infty$ for any fixed $x$: $$ \boxed{\Gamma(x)=\lim_{n\to\infty}\frac{n!\,n^x}{x(x+1)(x+2)\cdots(x+n)}} $$
Step 3 · The functional equation and $\Gamma(n+1)=n!$
Applying the limit formula at $x+1$ and multiplying numerator and denominator by $x$: $$ \Gamma(x+1)=\lim_{n\to\infty}\frac{n^{x+1}n!}{(x+1)(x+2)\cdots(x+n+1)} = x\lim_{n\to\infty}\frac{n^xn!}{x(x+1)\cdots(x+n)}\cdot\frac n{x+n+1}. $$ Since $\dfrac n{x+n+1}\to1$: $$ \boxed{\Gamma(x+1)=x\Gamma(x)} $$ At $x=1$: $\Gamma(1)=\lim_{n\to\infty}\dfrac{n\cdot n!}{(n+1)!}=\lim_{n\to\infty}\dfrac n{n+1}=1$. Iterating the functional equation, $\Gamma(n+1)=n\cdot(n-1)\cdots1\cdot\Gamma(1)=n!$: $$ \boxed{\Gamma(n+1)=n!\quad\text{for every integer }n\ge0.} $$
Summary Table
| Result | Formula |
|---|---|
| Euler's limit formula | $\Gamma(x)=\lim_{n\to\infty}\dfrac{n!\,n^x}{x(x+1)\cdots(x+n)}$ |
| Functional equation | $\Gamma(x+1)=x\Gamma(x)$ |
| $\Gamma(1)$ | $1$ |
| $\Gamma(n+1)$ | $n!$ |
Additional Notes
- Correction: a draft alternative derivation of $\Gamma(x+1)=x\Gamma(x)$ ("Method 2") claimed that $\prod_{n=1}^\infty\frac{n+1}{n+x+1}$ "telescopes to $x+1$" — this is false: at $x=1$ the partial products $\prod_{n=1}^N\frac{n+1}{n+2}=\frac2{N+2}\to0$, not to $x+1=2$. The valid derivation (Step 3 above, "Method 1" in the source) avoids this error entirely by working directly with Euler's limit formula and the elementary fact $n/(x+n+1)\to1$.
- This limit formula is the third of the three classical characterizations of $\Gamma$ encountered in this collection: Problem 387's Weierstrass infinite product, this problem's Euler limit, and the (unstated but standard) Euler integral $\Gamma(x)=\int_0^\infty t^{x-1}e^{-t}\,dt$ for $x>0$ — all provably equal, though only the first two are addressed here.
- The functional equation $\Gamma(x+1)=x\Gamma(x)$ is what makes $\Gamma$ the natural continuous interpolation of the factorial: it exactly mirrors the recursive definition $n!=n\cdot(n-1)!$, extended from integers to all $x\notin\{0,-1,-2,\ldots\}$.