I do not understand a claim in Emil Artin's book on the Gamma function. The claim is that $\theta$ is independent of $x$ and it occurs on page 22 of the English translation of the work.
I will try my best to explain the claim and steps that lead to it below.
For $x > 0$ define $$g(x) = (x+\frac{1}{2})\ln(1+\frac{1}{x})-1,$$ and $$\mu(x) = \sum_{n=0}^{\infty}g(x+n).$$
To see the series defining $\mu(x)$ converges pointwise note $$ \begin{align} g(x) &= \dfrac{2x+1}{2} \ln \dfrac{ 1 + \dfrac{1}{2x+1} }{1 - \dfrac{1}{2x+1}} - 1\\ &= \dfrac{1}{3(2x+1)^2}+\dfrac{1}{5(2x+1)^4}+\dots ( \text{ this is } > 0)\\ &<\dfrac{1}{3(2x+1)^2}+\dfrac{1}{3(2x+1)^4}+\dots \\ &< \dfrac{1}{3(2x+1)^2}\dfrac{1}{1-\dfrac{1}{(2x+1)^2}} = \dfrac{1}{12}(\dfrac{1}{x}-\dfrac{1}{x+1}) \end{align} $$
The above shows $0 < g(x) < \dfrac{1}{12}(\dfrac{1}{x} - \dfrac{1}{x+1})$ and also $$0 < \sum_{n=0}^{\infty}g(x+n) < \dfrac{1}{12x}$$
So $\mu(x)$ is well defined and $0 < \mu(x) < \dfrac{1}{12x}$. So far so good.
Next the claim follows, the claim is that $\mu(x) = \dfrac{\theta}{12x}$ where $0<\theta<1$ and $\theta$ is independent of $x$. Since no proof is given of this claim, it must be immediate, however I cannot see why this must be true.
Added later
$\theta$ is not independent of $x$, the claim is wrong. There is an expansion of $\mu(x)$ later in the text that makes it obvious.