TAOCP 1.2.7 Exercise 24

Section 1.2.7: Harmonic Numbers

Exercise 24. [HM21] Show that

$$ xe^{\gamma x} \prod_{k \ge 1} \left(\left(1+\frac{x}{k}\right)e^{-x/k}\right) = \frac{1}{\Gamma(x)}. $$

(Consider the partial products of this infinite product.)

Verified: yes
Solve time: 2m38s

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Solution

Let

$$ P_n(x)=x e^{\gamma x}\prod_{k=1}^{n}\left(1+\frac{x}{k}\right)e^{-x/k}. $$

We shall show that

$$ \lim_{n\to\infty}P_n(x)=\frac1{\Gamma(x)}. $$

The hint suggests working directly with the partial products.

Using $\Gamma(y+1)=y\Gamma(y)$, we have

$$ \Gamma(n+1+x) =\Gamma(x+1)\prod_{k=1}^{n}(x+k), $$

and

$$ \Gamma(n+1)=\prod_{k=1}^{n}k. $$

Therefore

$$ \frac{\Gamma(n+1)\Gamma(x+1)}{\Gamma(n+1+x)} = \prod_{k=1}^{n}\frac{k}{k+x} = \prod_{k=1}^{n}\left(1+\frac{x}{k}\right)^{-1}. $$

Hence

$$ \prod_{k=1}^{n}\left(1+\frac{x}{k}\right) = \frac{\Gamma(n+1+x)} {\Gamma(x+1)\Gamma(n+1)}. $$

Substituting this into $P_n(x)$ gives

$$ P_n(x) = \frac{x e^{\gamma x-xH_n}}{\Gamma(x+1)} \frac{\Gamma(n+1+x)}{\Gamma(n+1)}, $$

where

$$ H_n=\sum_{k=1}^{n}\frac1k. $$

Since $\Gamma(x+1)=x\Gamma(x)$,

$$ P_n(x) = \frac{e^{\gamma x-xH_n}}{\Gamma(x)} \frac{\Gamma(n+1+x)}{\Gamma(n+1)}. $$

Thus it remains to prove that

$$ e^{\gamma x-xH_n} \frac{\Gamma(n+1+x)}{\Gamma(n+1)} \longrightarrow 1. \tag{1} $$

From Exercise 23,

$$ H_n=\frac{\Gamma'(n+1)}{\Gamma(n+1)}+\gamma, $$

so

$$ \gamma-H_n=-\psi(n+1), $$

where

$$ \psi(z)=\frac{\Gamma'(z)}{\Gamma(z)}. $$

Taking logarithms of the expression in (1),

$$ L_n = \log\Gamma(n+1+x)-\log\Gamma(n+1) -x\psi(n+1). $$

Using

$$ \frac{d}{dt}\log\Gamma(t)=\psi(t), $$

we obtain

$$ L_n = \int_{n+1}^{,n+1+x}\psi(t),dt - x\psi(n+1) = \int_{0}^{x} \bigl(\psi(n+1+u)-\psi(n+1)\bigr),du . $$

Now

$$ \psi'(z) = \sum_{m=0}^{\infty}\frac1{(z+m)^2}, $$

hence for $z>0$,

$$ 0<\psi'(z) \le \sum_{m=0}^{\infty} \int_{m}^{m+1}\frac{dt}{(z+t)^2} = \int_{0}^{\infty}\frac{dt}{(z+t)^2} = \frac1z. $$

Therefore, by the mean value theorem,

$$ \bigl|\psi(n+1+u)-\psi(n+1)\bigr| \le u,\sup_{v\in[0,u]}\psi'(n+1+v) \le \frac{u}{n+1}. $$

Consequently,

$$ |L_n| \le \int_{0}^{x}\frac{u}{n+1},du = \frac{x^{2}}{2(n+1)} \longrightarrow 0. $$

Thus

$$ L_n\to0, $$

which implies

$$ e^{\gamma x-xH_n} \frac{\Gamma(n+1+x)}{\Gamma(n+1)} \to 1. $$

Returning to the expression for $P_n(x)$,

$$ P_n(x) = \frac1{\Gamma(x)} \left( e^{\gamma x-xH_n} \frac{\Gamma(n+1+x)}{\Gamma(n+1)} \right), $$

and the factor in parentheses tends to $1$. Hence

$$ \lim_{n\to\infty}P_n(x) = \frac1{\Gamma(x)}. $$

Since $P_n(x)$ is the $n$th partial product,

$$ x e^{\gamma x} \prod_{k\ge1} \left(\left(1+\frac{x}{k}\right)e^{-x/k}\right) = \frac1{\Gamma(x)}. $$

This is the desired identity. ∎