1. **Problem statement:** We want to show that there exists a convergent series of positive numbers $\sum_{n=1}^\infty a_n$ such that for any integer $k \geq 1$, there exists an index $n_k$ where the sequence $a_{n_k} < a_{n_k+1} < \cdots < a_{n_k+k}$ is strictly increasing. 2. **Key idea:** We need a convergent series with infinitely many arbitrarily long strictly increasing subsequences starting at some indices $n_k$. 3. **Construction:** Consider the series defined by blocks of increasing sequences with rapidly decreasing terms. For example, define for each $k \geq 1$ a block of length $k+1$: $$ a_{n_k+j} = \frac{j+1}{(k+1)!} \quad \text{for } j=0,1,\ldots,k $$ where $n_1=1$ and $n_{k+1} = n_k + (k+1)$. 4. **Check increasing property:** Within each block of length $k+1$, the terms are: $$ a_{n_k} = \frac{1}{(k+1)!} < \frac{2}{(k+1)!} < \cdots < \frac{k+1}{(k+1)!} = \frac{1}{k!} $$ which is strictly increasing. 5. **Check positivity:** All terms $a_n$ are positive since factorials and numerators are positive. 6. **Check convergence:** The sum of the series is $$ \sum_{k=1}^\infty \sum_{j=0}^k a_{n_k+j} = \sum_{k=1}^\infty \sum_{j=0}^k \frac{j+1}{(k+1)!} = \sum_{k=1}^\infty \frac{1}{(k+1)!} \sum_{j=0}^k (j+1) $$ The inner sum is an arithmetic series: $$ \sum_{j=0}^k (j+1) = \frac{(k+1)(k+2)}{2} $$ So the total sum is $$ \sum_{k=1}^\infty \frac{(k+1)(k+2)}{2 (k+1)!} = \sum_{k=1}^\infty \frac{k+2}{2 k!} $$ Since $k!$ grows faster than any polynomial, this series converges. 7. **Conclusion:** We have constructed a convergent series of positive terms with arbitrarily long strictly increasing subsequences starting at indices $n_k$ for each $k \geq 1$. **Final answer:** Such a series exists, for example, $$ a_n = \begin{cases} \frac{j+1}{(k+1)!} & \text{if } n = n_k + j, 0 \leq j \leq k \\ 0 & \text{otherwise} \end{cases} $$ with $n_1=1$ and $n_{k+1} = n_k + (k+1)$.