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I am triyng to solve the following problem in Brown and Churchill's complex variables textbook.

Show that the function $f_2 (z) = 1/z^2$ ($z \neq 0$) is the analytic continuation of the function \begin{align*} f_1 (z) = \sum\limits_{n=0}^{\infty} (n+1)(z + 1)^n \ \ \ (|z+1| < 1) \end{align*} into the domain consisting of all points in the $z$ plane except $z = 0$.

As a first note, I am having difficulty mapping the definition of analytic continuation to this problem. The definition in the textbook is that if we have two domains, say $D_1$ and $D_2$, where some function $f_1$ is analytic on $D_1$, some function $f_2$ is analytic on $D_2$, and $f_1 (z) = f_2 (z)$ on $D_1 \cap D_2$, where this intersection is nonempty, then $f_2$ is the analytic continuation of $f_1$ into $D_2$.

Assuming that I have not misstated that (please tell me if I have), we have: \begin{align*} D_1 = \{z \in \mathbb{C} : |z + 1| < 1\}, \ \ \ D_2 = \{z \in \mathbb{C} : z \neq 0\}. \end{align*} So we have \begin{align*} D_1 \cap D_2 = \{z \in \mathbb{C} : |z + 1| < 1 \text{ and } z \neq 0\}. \end{align*} From here, I am stuck. I know I need to prove that $\frac{1}{z^2} = \sum\limits_{n=0}^{\infty} (n+1)(z + 1)^n$ for any $z \in D_1 \cap D_2$. I don't know if I should try to demonstrate that the moduli are equal or expand $\frac{1}{z^2}$ in a power series and hope that these results will match, subject to the given constraint.

Any help would be greatly appreciated.

EDIT: I do not believe this question is a duplicate. I looked through the link below, and it does not address this problem, nor does it seem to deal with concepts in complex analysis.

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    $\begingroup$ Possible duplicate of How can I evaluate $\sum_{n=0}^\infty(n+1)x^n$? $\endgroup$ – Peter Foreman Aug 7 '19 at 19:57
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    $\begingroup$ Check again what definition you have for "continuation". It looks like you are done, you showed that the functions are equal on the intersection of the domains. The definition should say that this is the condition for a "continuation" to the greater domain. $\endgroup$ – Lutz Lehmann Aug 7 '19 at 20:20
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Since the geometric series $\sum_{n=0}^\infty (1+z)^{n+1}$ converges absolutely on $D_1$, we have $$ \sum_{n=0}^\infty(n+1)(1+z)^n = \frac{d}{dz}\left[\sum_{n=0}^\infty (1+z)^{n+1}\right] = \frac{d}{dz}\left[\frac{1}{1-(1+z)} - 1\right] = \frac{d}{dz}\left[-\frac{1}{z}\right] = \frac{1}{z^2}. $$

So $\sum_{n=0}^\infty (n+1)(1+z)^n = z^{-2}$ on all of $D_1$, and thus also on all of $D_1\cap D_2$.

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An echo-answer, confirming the computations done in the other two answers. Yes, on the smaller region, the geometric series can be summed to a rational function, which is $1/z^2$.

One should develop a sense that "we are done" at this point. Namely, at least for connected open sets in $\mathbb C$, (holomorphic) functions equal on a non-empty open are equal on the whole, by the identity principle...

True, when we talk about functions bureacratically defined only on some subsets of a larger region, it might not make bureaucratic sense to talk about their values outside the original "defined domains". But/and this is where the notion of analytic continuation is what we want, exactly.

In particular, one could easily inadvertently imagine that there is something more to show than agreement on a non-empty open... But by the identity principle, we are done.

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We need to show that $1/z^2$ is the analytic continuation of $S$, so amongst other things we need to show they are equivalent on $D_1\cap D_2$.

Splitting the sum,

\begin{eqnarray} S&=&\sum_{n=0}^\infty(z+1)^n+\sum_{n=0}^\infty n(z+1)^n\\ &=&\frac{1}{1-(z+1)}+(1+z)\frac{d}{dz}\sum_{n=0}^\infty (1+z)^n\\ &=&-\frac{1}{z}-(1+z)\frac{d}{dz}\frac{1}{z}\\ &=&-\frac{1}{z}+\frac{1+z}{z^2}\\ &=&\frac{1}{z^2}. \end{eqnarray}

Since $S$ and $1/z^2$ are both analytic functions in the domains $D_1$ and $D_2$ respectively, and the intersection $D_1\cap D_2\neq\emptyset$, and furthermore $S=1/z^2$ on $D_1\cap D_2$, as shown above, then $1/z^2$ is the analytic continuation of $S$ to $\mathbb{C}\setminus\{0\}$, and vice-versa. Furthermore, this analytic continuation from $S$ to $1/z^2$ is unique.

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