Let $\mathcal{F}$ be the family of all nonempty finite subsets of $\mathbb{N} \cup \{0\}.$ Find all real numbers $a$ for which the series
$$\sum_{A \in \mathcal{F}} \frac{1}{\sum_{k \in A}a^k}$$
is convergent.
1. **Case \( a \leq 1 \):**
   - For \( a \leq 1 \), consider the series:
     \[
     \sum_{A \in \mathcal{F}} \frac{1}{\sum_{k \in A} a^k}
     \]
   - Since \( a \leq 1 \), each term \( a^k \leq 1 \). Therefore, for any nonempty finite subset \( A \subseteq \mathbb{N} \cup \{0\} \), we have:
     \[
     \sum_{k \in A} a^k \leq |A|
     \]
   - This implies:
     \[
     \frac{1}{\sum_{k \in A} a^k} \geq \frac{1}{|A|}
     \]
   - The series:
     \[
     \sum_{A \in \mathcal{F}} \frac{1}{|A|}
     \]
     diverges because it includes the harmonic series as a subseries. Hence, the original series diverges for \( a \leq 1 \).

2. **Case \( a > 1 \):**
   - Let \( a > 1 \). Denote by \( S_n \) the partial sum running through the subsets whose elements do not exceed \( n \):
     \[
     S_n = \sum_{A \subseteq \{0, 1, \ldots, n\}} \frac{1}{\sum_{k \in A} a^k}
     \]
   - We can write:
     \[
     S_n = S_{n-1} + \sum_{A \subseteq \{0, 1, \ldots, n-1\}} \frac{1}{a^n + \sum_{k \in A} a^k}
     \]
   - The latter sum consists of \( 2^n \) summands. Each summand is bounded above by:
     \[
     \frac{1}{a^n}
     \]
     and bounded below by:
     \[
     \frac{1}{\sum_{k=0}^n a^k} = \frac{a-1}{a^{n+1} - 1}
     \]
   - Therefore:
     \[
     S_n \leq S_{n-1} + 2^n \cdot \frac{1}{a^n}
     \]
     \[
     S_n \leq S_{n-1} + \left(\frac{2}{a}\right)^n
     \]
   - Summing this inequality, we get:
     \[
     S_n \leq \sum_{j=0}^n \left(\frac{2}{a}\right)^j
     \]
   - The series \( \sum_{j=0}^\infty \left(\frac{2}{a}\right)^j \) converges if and only if \( \frac{2}{a} < 1 \), i.e., \( a > 2 \).

3. **Lower Bound Analysis:**
   - For \( a \leq 2 \), consider:
     \[
     S_n \geq \sum_{A \subseteq \{0, 1, \ldots, n-1\}} \frac{1}{a^n + \sum_{k \in A} a^k}
     \]
   - Each term is at least:
     \[
     \frac{1}{a^n + \sum_{k=0}^{n-1} a^k} = \frac{1}{a^n + \frac{a^n - 1}{a-1}} = \frac{a-1}{a^n (a-1) + a^n - 1} \approx \frac{a-1}{a^{n+1}}
     \]
   - Thus:
     \[
     S_n \geq \sum_{j=0}^n \frac{2^n (a-1)}{a^{n+1}} = \frac{a-1}{a^{n+1}} \cdot 2^n
     \]
   - For \( a \leq 2 \), this lower bound diverges as \( n \to \infty \).

Conclusion:
- The series converges if and only if \( a > 2 \).

The final answer is \( \boxed{ a > 2 } \)