Moving Center of Power Series Given a power series:
$$\lim_{N\to\infty}\sum_{k=0}^N A_k (z-a)^k$$
I expand the powers:
$$\lim_{N\to\infty}\sum_{l=0}^N(\sum_{k=l}^N A_k \binom{k}{l}(-1)^{k-l}a^{k-l})z^l$$
But here I face the problem that the coefficients now depend on the limit, so I cannot directly apply the root criterion. I could take the limit of the coefficients first and then apply the root criterion. However that would correspond to a double limit. So how can I solve this problem instead?
 A: Writing your series with a limit is a bit confusing. However, that aside, you then formulate a limit of the partial sums of the shifted partial sums which is not a shifted power series expansion for the original function (although it seems the limit exists and equals your function). In general, taking a partial sum and re-writing to match another "center" may not necessarily yield a partial sum for that other center. 
I don't think this is what you intended. Let's be clear by naming things, $$f(z)=\lim_{n\to\infty}\sum_{k=0}^n A_k (z-a)^k.$$ This function, $f$, is called a series centered at $a$. By Taylor's theorem $A_k=\frac{f^{(k)}(a)}{k!}$ are the series coefficients. To relocate the center of the series, one would recalculate the series coefficients, for example, lets rewrite $f$ so that it is centered at zero,  $$f(z)=\lim_{n\to\infty}\sum_{k=0}^n \frac{f^{(k)}(0)}{k!}z^k.$$ If you want to use your first representation to calculate the coefficients of the second representation, then you would have the following, 
$$f(z)=\lim_{n\to\infty} \sum_{k=0}^n \frac{1}{k!} \left(\left.\frac{d^k}{dz^k}\left(\lim_{m\to\infty}\sum_{j=0}^m A_j (z-a)^j\right) \right|_{z=0}\right) z^k. $$ 
Which gives, 
$$f(z)=\lim_{n\to\infty} \sum_{k=0}^n \frac{1}{k!} \left( \lim_{m\to\infty}\sum_{j=0}^m k!\binom{j}{k} A_j (-a)^{j-k}\right) z^k $$ $$=\lim_{n\to\infty} \sum_{k=0}^n \left( \lim_{m\to\infty}\sum_{j=0}^m \binom{j}{k} A_j (-a)^{j-k}\right) z^k. $$ What you wrote, is an expanded version of one of the partial sums in the original series. However, that is not equal to the partial sums of a re-centered power series. I think you wanted the formula with two limits anyway, hence I think your problem is solved.   
Hope this makes sense. 
