$A$ is a positive real number.$n$ is positive integer number.Find the set of possible values of the infinite sum $x_0^n+x_1^n+x_2^n+...$ where $x_0,x_1,x_2...$ are all positive real numbers so that the infinite series $x_0+x_1+x_2+...$ has sum $A$.
1. **Define the normalized variables:**
   Let \( y_i = \frac{x_i}{A} \). Then, since \( x_0 + x_1 + x_2 + \dots = A \), we have:
   \[
   y_0 + y_1 + y_2 + \dots = 1
   \]
   Since \( y_i \) are positive real numbers and their sum is 1, it follows that \( 0 \le y_i \le 1 \).

2. **Analyze the sum of powers:**
   Consider the sum \( x_0^n + x_1^n + x_2^n + \dots \). Using the normalized variables, we can write:
   \[
   x_i = A y_i \implies x_i^n = (A y_i)^n = A^n y_i^n
   \]
   Therefore, the sum becomes:
   \[
   x_0^n + x_1^n + x_2^n + \dots = A^n (y_0^n + y_1^n + y_2^n + \dots)
   \]

3. **Establish the upper bound:**
   Since \( 0 \le y_i \le 1 \), it follows that \( y_i^n \le y_i \). Thus:
   \[
   y_0^n + y_1^n + y_2^n + \dots \le y_0 + y_1 + y_2 + \dots = 1
   \]
   Hence:
   \[
   A^n (y_0^n + y_1^n + y_2^n + \dots) \le A^n \cdot 1 = A^n
   \]
   Equality holds when \( y_0 = 1 \) and \( y_1 = y_2 = \dots = 0 \), which corresponds to \( x_0 = A \) and \( x_1 = x_2 = \dots = 0 \). Therefore, the maximum value of the sum is \( A^n \).

4. **Establish the lower bound:**
   To find the lower bound, consider the case where \( x_i = \frac{A}{k} \) for \( i \le k-1 \) and \( x_i = 0 \) for \( i \ge k \). Then:
   \[
   x_0^n + x_1^n + \dots + x_{k-1}^n = k \left( \frac{A}{k} \right)^n = k \cdot \frac{A^n}{k^n} = \frac{A^n}{k^{n-1}}
   \]
   As \( k \to \infty \), \( \frac{A^n}{k^{n-1}} \to 0 \). Therefore, arbitrarily small values can be obtained.

5. **Intermediate values:**
   By continuity, every intermediate value between 0 and \( A^n \) can be attained. Thus, the set of possible values is the interval \( (0, A^n] \).

6. **Special case for \( n = 1 \):**
   For \( n = 1 \), the sum \( x_0 + x_1 + x_2 + \dots = A \) is fixed. Therefore, the set of possible values is just \( \{A\} \).

The final answer is \( \boxed{ (0, A^n] } \) for \( n > 1 \) and \( \{A\} \) for \( n = 1 \).