Consider the following sequence
$$(a_n)_{n=1}^{\infty}=(1,1,2,1,2,3,1,2,3,4,1,2,3,4,5,1,\dots)$$
Find all pairs $(\alpha, \beta)$ of positive real numbers such that $\lim_{n\to \infty}\frac{\displaystyle\sum_{k=1}^n a_k}{n^{\alpha}}=\beta$.

(Proposed by Tomas Barta, Charles University, Prague)
1. **Identify the structure of the sequence \( (a_n) \):**
   The sequence \( (a_n) \) is constructed by concatenating increasing sequences of natural numbers. Specifically, the sequence starts with \(1\), then \(1, 2\), then \(1, 2, 3\), and so on. The \(n\)-th segment of the sequence is \(1, 2, \ldots, n\).

2. **Determine the position of the \(n\)-th term:**
   The total number of terms up to the \(n\)-th segment is given by the sum of the first \(n\) natural numbers:
   \[
   T(n) = \frac{n(n+1)}{2}
   \]
   This means that the \(T(n)\)-th term is the last term of the \(n\)-th segment.

3. **Sum of the sequence up to \(T(n)\):**
   The sum of the sequence up to \(T(n)\) can be calculated by summing the sums of the first \(n\) segments:
   \[
   \sum_{k=1}^{T(n)} a_k = \sum_{k=1}^{n} \sum_{j=1}^{k} j = \sum_{k=1}^{n} \frac{k(k+1)}{2} = \frac{1}{2} \sum_{k=1}^{n} k^2 + \frac{1}{2} \sum_{k=1}^{n} k
   \]
   Using the formulas for the sum of the first \(n\) natural numbers and the sum of the squares of the first \(n\) natural numbers:
   \[
   \sum_{k=1}^{n} k = \frac{n(n+1)}{2}
   \]
   \[
   \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}
   \]
   Therefore:
   \[
   \sum_{k=1}^{T(n)} a_k = \frac{1}{2} \cdot \frac{n(n+1)(2n+1)}{6} + \frac{1}{2} \cdot \frac{n(n+1)}{2} = \frac{n(n+1)(2n+1)}{12} + \frac{n(n+1)}{4}
   \]
   Simplifying the expression:
   \[
   \sum_{k=1}^{T(n)} a_k = \frac{n(n+1)}{12} (2n+1) + \frac{3n(n+1)}{12} = \frac{n(n+1)}{12} (2n+1 + 3) = \frac{n(n+1)(2n+4)}{12} = \frac{n(n+1)(n+2)}{6}
   \]

4. **Asymptotic behavior of the sum:**
   For large \(n\), the dominant term in \(\frac{n(n+1)(n+2)}{6}\) is \(\frac{n^3}{6}\). Thus:
   \[
   \sum_{k=1}^{T(n)} a_k \sim \frac{n^3}{6}
   \]

5. **Analyzing the limit:**
   We need to find \(\alpha\) and \(\beta\) such that:
   \[
   \lim_{n \to \infty} \frac{\sum_{k=1}^n a_k}{n^\alpha} = \beta
   \]
   Using the asymptotic behavior, we have:
   \[
   \frac{\sum_{k=1}^{T(n)} a_k}{[T(n)]^\alpha} \sim \frac{\frac{n^3}{6}}{\left(\frac{n(n+1)}{2}\right)^\alpha} \sim \frac{\frac{n^3}{6}}{\left(\frac{n^2}{2}\right)^\alpha} = \frac{\frac{n^3}{6}}{\frac{n^{2\alpha}}{2^\alpha}} = \frac{2^\alpha n^{3-2\alpha}}{6}
   \]
   For the limit to be finite and non-zero, the exponent of \(n\) must be zero:
   \[
   3 - 2\alpha = 0 \implies \alpha = \frac{3}{2}
   \]
   Substituting \(\alpha = \frac{3}{2}\):
   \[
   \frac{2^{3/2} n^{3-2(3/2)}}{6} = \frac{2^{3/2}}{6} = \frac{\sqrt{8}}{6} = \frac{2\sqrt{2}}{6} = \frac{\sqrt{2}}{3}
   \]
   Thus, \(\beta = \frac{\sqrt{2}}{3}\).

Conclusion:
\[
(\alpha, \beta) = \left(\frac{3}{2}, \frac{\sqrt{2}}{3}\right)
\]

The final answer is \( \boxed{ \left(\frac{3}{2}, \frac{\sqrt{2}}{3}\right) } \)