How to prove that $e^{\gamma}={e^{H_{x-1}}\over x}\prod_{n=0}^{\infty}\cdots?$

How do we show that $$e^{\gamma}={e^{H_{x-1}}\over x}\prod_{n=0}^{\infty}\left(\prod_{k=0}^{n}\left({k+x+1\over k+x}\right)^{(-1)^{k}{{n\choose k}}}\right)^{1\over n+2}?\tag1$$

Where $\gamma$ is the Euler-Mascheroni constant

$H_0=0$, $H_n$ is the harmonic number.

$x\ge1$

Take the log of $(1)$

$$\gamma-H_{x-1}+\ln(x)=\sum_{n=0}^{\infty}{1\over n+2}\sum_{k=0}^{n}(-1)^k{n\choose k}\ln\left({k+x+1\over k+x}\right)\tag2$$

Probably take $(2)$ into an integral and take it from there...?

• looks like euler acceleration – tired Sep 28 '17 at 21:01
• How does $(-1)^{\binom{n}{k}} = (-1)^k \binom{n}{k}$? – Runemoro Sep 28 '17 at 22:44

This can indeed be solved by Euler series acceleration. Notice first that

$$f(s) := \log\left(1 + \frac{1}{x+s}\right) = \int_{0}^{\infty} \frac{1 - e^{-u}}{u} \, e^{-su} \, du.$$

Then its $n$-fold forward difference satisfies

$$(-1)^n \Delta^n f (s) = \sum_{k=0}^{n} (-1)^{k}\binom{n}{k} f(s+k) = \int_{0}^{\infty} \frac{(1 - e^{-u})^{n+1}}{u}\, e^{-su} \, du, \tag{1}$$

which is monotone decreasing to $0$ as $s \to \infty$. Next we refer to the following version of the Euler series acceleration: (See my blog posting for a proof, for instance.)

Theorem. If $\sum_{n=0}^{\infty} a_n z^n$ has radius of convergence $\geq 1$. Then for any $t \in [0, 1)$,

$$\sum_{n=0}^{\infty} (-1)^n a_n t^n = \frac{1}{1+t} \sum_{n=0}^{\infty} (-1)^n (\Delta^n a)_0 \left( \frac{t}{1+t}\right)^n$$

holds.

In order to apply this to our case, set $z = t/(1+t)$ and plug $a_n = f(x+n)$, where $x > 0$. Then for $t \in [0, 1)$ it follows that

\begin{align*} \sum_{n=0}^{\infty} (-1)^n (\Delta^n a)_0 z^{n+1} &= \sum_{n=0}^{\infty} (-1)^n a_n t^{n+1} \\ &= \sum_{n=0}^{\infty} (-1)^n t^{n+1} \int_{0}^{\infty} \frac{1 - e^{-u}}{u} \, e^{-(x+n)u} \, du \\ &= \int_{0}^{\infty} \frac{1 - e^{-u}}{u} \frac{t e^{-xu}}{1 + t e^{-u}} \, du \\ &= \int_{0}^{\infty} \frac{z (1 - e^{-u}) e^{-xu}}{1 - z(1 - e^{-u})} \, \frac{du}{u}. \end{align*}

Although this manipulation holds only when $z \in [0, \frac{1}{2})$ (mainly because of the restriction of the theorem above), now both sides define a holomorphic function on $\mathbb{D} = \{z \in \mathbb{C} : |z| < 1\}$ and hence the identity extends to all of $\mathbb{D}$ by the principle of analytic continuation.

Moreover, the left-hand side converges even when $z = 1$ due to the alternating series test combined with our previous remark on $\text{(1)}$. So by the Abel's theorem,

\begin{align*} S(x) &:= \sum_{n=0}^{\infty} \frac{1}{n+2} \sum_{k=0}^{n} (-1)^k \binom{n}{k} \log\left(1+\frac{1}{k+x}\right) \\ &= \sum_{n=0}^{\infty} (-1)^n (\Delta^n a)_0 \int_{0}^{1} z^{n+1} \, dz \\ &= \int_{0}^{1} \sum_{n=0}^{\infty} (-1)^n (\Delta^n a)_0 z^{n+1} \, dz \\ &= \int_{0}^{1} \left( \int_{0}^{\infty} \frac{z (1 - e^{-u}) e^{-xu}}{1 - z(1 - e^{-u})} \, \frac{du}{u} \right) \, dz \\ &= \int_{0}^{\infty} \left( \frac{1}{1 - e^{-u}} - \frac{1}{u} \right) e^{-xu} \, du. \end{align*}

Computing this integral is not hard; differentiate both sides to obtain $S'(x) = \frac{1}{x} - \psi'(x)$ and then utilize the condition $\lim_{x\to\infty}S(x) = 0$ to obtain

$$S(x) = \log x - \psi (x) = \log x + \gamma - H_{x-1}.$$

• pretty nice (+1) – tired Oct 2 '17 at 13:27

Here's a possible start manipulating the double sum.

\begin{align}\\ S(x)&= \sum_{n=0}^{\infty}{1\over n+2}\sum_{k=0}^{n}(-1)^k{n\choose k}\ln\left({k+x+1\over k+x}\right)\\ &=\sum_{n=0}^{\infty}{1\over n+2} (\sum_{k=0}^{n}(-1)^k{n\choose k}\ln(k+x+1))-\sum_{k=0}^{n}(-1)^k{n\choose k}\ln(k+x)))\\ &=\sum_{n=0}^{\infty}{1\over n+2} (\sum_{k=1}^{n+1}(-1)^{k-1}{n\choose k-1}\ln(k+x)-\sum_{k=0}^{n}(-1)^k{n\choose k}\ln(k+x))\\ &=\sum_{n=0}^{\infty}{1\over n+2} (\sum_{k=1}^{n}\ln(k+x)((-1)^{k-1}{n\choose k-1}-(-1)^k{n\choose k})\\ &\quad +(-1)^n{n \choose n}\ln(n+1+x)-(-1)^0{n \choose 0}\ln(x) )\\ &=\sum_{n=0}^{\infty}{1\over n+2} (\sum_{k=1}^{n}\ln(k+x)(-1)^{k-1}({n\choose k-1}+{n\choose k})\\ &\quad +(-1)^n\ln(n+1+x)-\ln(x) )\\ &=\sum_{n=0}^{\infty}{1\over n+2} (\sum_{k=1}^{n}\ln(k+x)(-1)^{k-1}{n+1\choose k}\\ &\quad +(-1)^n\ln(n+1+x)-\ln(x) )\\ &=\sum_{n=0}^{\infty}{1\over n+2} \sum_{k=0}^{n+1}\ln(k+x)(-1)^{k-1}{n+1\choose k}\\ &=-\sum_{n=0}^{\infty}{1\over n+2} \Delta^{n+1}\ln(x)\\ \end{align}

This almost looks like a Newton series.

Not sure where to go from here, so I'll leave it at this in the hope that someone else can take it further.