1. **Base Case: $n = 2$**
- For $n = 2$, we need to show that $\left| \sum_{k=1}^{2} a_k - \sum_{k=1}^{2} A_k \right| < \frac{1}{2}$.
- We have $A_1 = a_1$ and $A_2 = \frac{a_1 + a_2}{2}$.
- Therefore, $\sum_{k=1}^{2} A_k = a_1 + \frac{a_1 + a_2}{2} = \frac{2a_1 + a_2}{2}$.
- Also, $\sum_{k=1}^{2} a_k = a_1 + a_2$.
- Thus, $\left| \sum_{k=1}^{2} a_k - \sum_{k=1}^{2} A_k \right| = \left| (a_1 + a_2) - \frac{2a_1 + a_2}{2} \right| = \left| a_1 + a_2 - a_1 - \frac{a_2}{2} \right| = \left| \frac{a_2}{2} \right|$.
- Since $a_2 \leq 1$, we have $\left| \frac{a_2}{2} \right| \leq \frac{1}{2}$.
- Therefore, $\left| \sum_{k=1}^{2} a_k - \sum_{k=1}^{2} A_k \right| < \frac{1}{2}$.

2. **Inductive Step: Assume the statement is true for $n = k$**
- Assume $\left| \sum_{i=1}^{k} a_i - \sum_{i=1}^{k} A_i \right| < \frac{k-1}{2}$ holds for some $k \geq 2$.

3. **Consider $n = k+1$**
- We need to show that $\left| \sum_{i=1}^{k+1} a_i - \sum_{i=1}^{k+1} A_i \right| < \frac{k}{2}$.
- We have $\sum_{i=1}^{k+1} a_i = \sum_{i=1}^{k} a_i + a_{k+1}$.
- Also, $\sum_{i=1}^{k+1} A_i = \sum_{i=1}^{k} A_i + A_{k+1}$, where $A_{k+1} = \frac{\sum_{i=1}^{k+1} a_i}{k+1}$.
- Therefore, $\left| \sum_{i=1}^{k+1} a_i - \sum_{i=1}^{k+1} A_i \right| = \left| \sum_{i=1}^{k} a_i + a_{k+1} - \left( \sum_{i=1}^{k} A_i + \frac{\sum_{i=1}^{k+1} a_i}{k+1} \right) \right|$.

4. Simplify the expression
- $\left| \sum_{i=1}^{k+1} a_i - \sum_{i=1}^{k+1} A_i \right| = \left| \sum_{i=1}^{k} a_i + a_{k+1} - \sum_{i=1}^{k} A_i - \frac{\sum_{i=1}^{k} a_i + a_{k+1}}{k+1} \right|$.
- Let $S_k = \sum_{i=1}^{k} a_i$, then the expression becomes $\left| S_k + a_{k+1} - \sum_{i=1}^{k} A_i - \frac{S_k + a_{k+1}}{k+1} \right|$.
- This simplifies to $\left| S_k + a_{k+1} - \sum_{i=1}^{k} A_i - \frac{S_k}{k+1} - \frac{a_{k+1}}{k+1} \right|$.
- Further simplifying, we get $\left| S_k + a_{k+1} - \sum_{i=1}^{k} A_i - \frac{S_k}{k+1} - \frac{a_{k+1}}{k+1} \right| = \left| S_k - \sum_{i=1}^{k} A_i + a_{k+1} - \frac{S_k}{k+1} - \frac{a_{k+1}}{k+1} \right|$.
- This can be written as $\left| S_k - \sum_{i=1}^{k} A_i - \frac{S_k}{k+1} + a_{k+1} - \frac{a_{k+1}}{k+1} \right|$.

5. Use the inductive hypothesis
- By the inductive hypothesis, $\left| S_k - \sum_{i=1}^{k} A_i \right| < \frac{k-1}{2}$.
- We need to show that $\left| S_k - \sum_{i=1}^{k} A_i - \frac{S_k}{k+1} + a_{k+1} - \frac{a_{k+1}}{k+1} \right| < \frac{k}{2}$.

6. Bound the additional terms
- Note that $\left| \frac{S_k}{k+1} \right| \leq \frac{k}{k+1}$ since $S_k \leq k$ and $\left| \frac{a_{k+1}}{k+1} \right| \leq \frac{1}{k+1}$.
- Therefore, $\left| \frac{S_k}{k+1} + \frac{a_{k+1}}{k+1} \right| \leq \frac{k}{k+1} + \frac{1}{k+1} = 1$.

7. Combine the bounds
- We have $\left| S_k - \sum_{i=1}^{k} A_i - \frac{S_k}{k+1} + a_{k+1} - \frac{a_{k+1}}{k+1} \right| \leq \left| S_k - \sum_{i=1}^{k} A_i \right| + \left| \frac{S_k}{k+1} + \frac{a_{k+1}}{k+1} \right|$.
- Using the inductive hypothesis and the bound on the additional terms, we get $\left| S_k - \sum_{i=1}^{k} A_i \right| + 1 < \frac{k-1}{2} + 1 = \frac{k+1}{2}$.

8. Conclusion
- Therefore, $\left| \sum_{i=1}^{k+1} a_i - \sum_{i=1}^{k+1} A_i \right| < \frac{k}{2}$.

By induction, the statement is true for all $n \geq 2$.

$\blacksquare$