I want to show that (exercise 8 in Amann & Escher Analysis 2):
\begin{equation} \log(\Gamma)(1+x) = -Cx + \sum_{k=2}^\infty (-1)^k \frac{\zeta(k)}{k} x^k \end{equation} for $x \in (-1, 1)$, where $C$ is the Euler-Mascheroni constant.
I tried to use the fact that: \begin{equation} \frac{1}{\Gamma(z)} = ze^{Cz}\prod_{k=1}^\infty(1+\frac{z}{k})e^{-z/k} \end{equation} I then tried this: \begin{equation} \log(1/\Gamma(z)) = -\log(\Gamma(z))= \log(z) + Cz + \sum_{k=1}^\infty (\log(1 + z/k) -z/k) = \\ \log(z) + Cz + \sum_{k=1}^\infty (\sum_{i=1}^\infty \frac{(-1)^{i-1}}{i} (\frac{z}{k})^i) - z/k = \log(z) + Cz + \sum_{k=1}^\infty \sum_{i=2}^\infty \frac{(-1)^{i-1}}{i} (\frac{z}{k})^i \end{equation} Hence, I have: \begin{equation} \log(\Gamma(z)) = -\log(z) - Cz + \sum_{k=1}^\infty \sum_{i=2}^\infty \frac{(-1)^{i}}{i} (\frac{z}{k})^i = -\log(z) - Cz + \sum_{i=2}^\infty \frac{(-1)^{i}}{i} \sum_{k=1}^\infty (\frac{z}{k})^i = \\ -\log(z) - Cz + \sum_{i=2}^{\infty} \frac{(-1)^{i} z^i }{i} \cdot \zeta(i) \end{equation} or, alternatively: \begin{equation} \log(\Gamma(1+x)) = -\log(1+x) - C(1+x) + \sum_{i=2}^{\infty} \frac{(-1)^{i} (1+x)^i }{i} \cdot \zeta(i) \end{equation}
And that is where I am stuck. I don't see how I can get rid of $(1+x)^i$ and arrive at $x^i$ in the numerator of the last formula. I am not even sure that the step with the change of summation order that I did can be applied here, given that for $x \in (0, 1)$ we have $|x + 1| > 1$, hence, the series $\sum_{i=2}^\infty \frac{(-1)^{i-1}}{i} (\frac{1+x}{k})^i$ is not converging absolutely. Could someone please help me?