Equivalence of holomorphic functions 
Given that $$\left(1-\frac{z}{\zeta_j}\right)^{-z}=\sum\limits_{k=1}^\chi\frac{z^k}{k\zeta_j^k},$$ where $\chi$ is the largest nonnegative integer $k$ for which $\sum\limits_{j=1}^\infty\frac{1}{|\zeta_j|^k}$ diverges. For what value of $\zeta_j$ is the aforementioned equality true?

Reduced Form: Observe that if $r=\frac{z}{\zeta_j}$, then the statement of the problem is given by $$\left(1-r\right)^{-z}=\sum\limits_{k=1}^\chi\frac{r^k}{k}.$$ Thus, by means of integration and substitution the equality becomes:
$$\left(\frac{1}{1-\frac{z}{\zeta_j}}\right)^{z}=\sum\limits_{k=1}^{\chi+1}\binom {\chi+1} k\frac{(-1)^k}{k}\left(1-\frac{z}{\zeta_j}\right)^z$$
Recognizing the left hand side as a geometric series to the power of $z$ and manipulating the expression yields:
$$\left(\frac{1}{1-(\frac{z}{\zeta_j})^{\chi+1}}\frac{1-(\frac{z}{\zeta_j})^{\chi+1}}{1-\frac{z}{\zeta_j}}\right)^{z}=\left(\frac{1}{1-(\frac{z}{\zeta_j})^{\chi+1}}\sum\limits_{k=1}^{\chi+1}\left(\frac{z}{\zeta_j}\right)^k\right)^{z}=\left(\sum\limits_{k=1}^{\chi+1}\frac{(\frac{z}{\zeta_j})^k}{1-(\frac{z}{\zeta_j})^{\chi+1}}\right)^z$$ It now follows that:
$$\left(\sum\limits_{k=1}^{\chi+1}\frac{(\frac{z}{\zeta_j})^k}{1-(\frac{z}{\zeta_j})^{\chi+1}}\right)^z=\sum\limits_{k=1}^{\chi+1}\binom {\chi+1} k\frac{(-1)^k}{k}\left(1-\frac{z}{\zeta_j}\right)^z$$ Taking the natural logarithm of both sides and hence exponentiating results in:
$$\sum\limits_{k=1}^{\chi+1}\frac{e^z(\frac{z}{\zeta_j})^k}{1-(\frac{z}{\zeta_j})^{\chi+1}}=\sum\limits_{k=1}^{\chi+1}\binom {\chi+1} k\frac{(-1)^k}{k}\left(1-\frac{z}{\zeta_j}\right)^z$$ Therefore, the form is now "reduced", whereby the series have equivalent indices.
Best Regards
 A: First, note that the index $j$ plays no role here, so let us forget about it. Now, your question is slightly ambiguous, I understand it as "if $D$ is the intersection of the domains of definition of all the $S_n$, find $\zeta$ such that $S_n (z) \to 1 \; \forall z \in D$". If the title has any value, then you would want this convergence to also be uniform.
Please note that $z=0$ is in the domain of convergence of all the $S_n$, therefore $z \in D$. But $S_n (z) = 0$, so you cannot have $S_n \to 1$, therefore your problem as stated now has no solution.
Maybe my understanding is mistaken and you meant something different, in which case I suggest reformulating your question.
In any case, if my forgotten elementary complex analysis still serves, $\lim \limits _{n \to \infty} S_n (z) = - \ln \big( 1- \frac z \zeta \big) - \Bbb e ^{-z \ln \big( 1- \frac z \zeta \big)}$, with all the precautions taken concerning the domain of definition of the logarithm. Therefore, you seem to ask for $\zeta$ such that $- \ln \big( 1- \frac z \zeta \big) - \Bbb e ^{-z \ln \big( 1- \frac z \zeta \big)} = 1 \; \forall z \in D$. Besides the mistake pointed above, I fail to see how a complicated holomorphic function such as the one on the left-hand side could be constant. With a correct statement of the problem I might try to equate the coefficient of $z$ in the Taylor series (i.e. the derivative) to $0$, impose that to be true $\forall z$ and see if this gives me anything (I doubt).
