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.


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...?

  • 1
    $\begingroup$ looks like euler acceleration $\endgroup$ – tired Sep 28 '17 at 21:01
  • $\begingroup$ How does $(-1)^{\binom{n}{k}} = (-1)^k \binom{n}{k}$? $\endgroup$ – 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 $t \in [0, 1)$,

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


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

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

  • $\begingroup$ pretty nice (+1) $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.