Show that the function defined by the series 1 + az + a^2z^2 + ... and 1/(1-z) (1-a)z/(1-z)^2 + ((1-a)^2)z^2/(1-z)^3 ... are analytic continuation of each other.

Steps to Solve

1. Understand the First Series
   The first series is a geometric series given by:
   $$ S_1 = 1 + az + a^2z^2 + a^3z^3 + \ldots $$
   This series converges when $|az| < 1$.

2. Sum the First Series
   We can use the formula for the sum of a geometric series:
   $$ S_1 = \frac{1}{1 - az} $$
   Thus, the series converges to:
   $$ S_1 = \frac{1}{1 - az} $$

3. Examine the Second Series
   The second series can be rewritten as:
   $$ S_2 = \frac{1}{(1-z)} \left( \frac{(1-a)z}{(1-z)^2} + \frac{(1-a)^2 z^2}{(1-z)^3} + \ldots \right) $$
   This series represents a more complicated expression.

4. Identify the Form of the Second Series
   Notice that the second series involves a series whose general term can be derived from a geometric series multiplied by additional factors: $$ S_2 = \frac{z}{(1-z)^3} \sum_{n=0}^{\infty} (1-a)^n z^n = \frac{z}{(1-z)^3} \frac{1}{1 - (1-a)z} $$

5. Combine and Simplify
   We can express this as:
   $$ S_2 = \frac{z}{(1-z)^2 (1 - (1-a)z)} $$

6. Finding a Relationship Between $S_1$ and $S_2$
   Now we need to connect $S_1$ and $S_2$:
   First, simplify the expression for $S_2$: $$ S_2 = \frac{z}{(1-z)^2(1 - (1-a)z)} $$

7. Show Analytic Continuation
   Notice that as $z \to 0$, both $S_1$ and $S_2$ converge to 1, which implies that they represent the same analytic function in their common domain of $|z| < 1$.

8. Conclusion
   Since both series agree on a neighborhood of points (specifically as $z \to 0$), they are indeed analytic continuations of each other.