How to determine summability of "nested series"

Some quick things one can notice:

• Your sum is bounded below by $$\sum_{k=1}^\infty a_k b_1$$ which, if the sum of $a_k$ is not convergent, is not convergent.
• If all three sums are convergent, then your sum is also convergent, because, if $$\sum_{j=1}^\infty b_j=B, \sum_{i=1}^\infty c_i=C,$$ then $$\sum_{k = 1}^\infty a_k \sum_{j = 1}^k b_j \sum_{i = j}^k c_i\leq \sum_{k=1}^\infty a_k\cdot B\cdot C$$

That covers 5 of the 8 cases.

The other three cases, I think, can swing either way, depending on how quickly one or the other series diverges.

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The most general result

The property that might be most useful to you is this:

If $a_k$ are nonnegative and $B_k$ is a non-decreasing positive sequence, then:

• If $$\sum_{k=1}^\infty a_k$$ converges and $B_k$ converges, then $$\sum_{k=1}^\infty a_k\cdot B_k$$ converges
• If $$\sum_{k=1}^\infty a_k$$ diverges, then $$\sum_{k=1}^\infty a_k\cdot B_k$$ diverges
• If $$\sum_{k=1}^\infty a_k$$ converges and $B_k$ diverges, then it is possible for the sum $$\sum_{k=1}^\infty a_k\cdot B_k$$ to both diverge and converge.

The three points above can be quickly proven.

• If $$\sum_{k=1}^\infty a_k$$ converges and $B_k$ converges, then $B_k$ has a limit $B$ which is also its upper limit, so $$\sum_{k=1}^\infty a_k B_k < \sum_{k=1}^\infty a_k B < \infty$$ so the sum converges.
• If $$\sum_{k=1}^\infty a_k$$ diverges, then $$\sum_{k=1}^\infty a_k\cdot B_k > \sum_{k=1}^\infty a_k\cdot B_1 = B_1\cdot\sum_{k=1}^\infty a_k = \infty$$ so the sum diverges.
• Take $a_k = \alpha^{-k}$ and define $B_k = \beta^k$. Then, clearly, $$\sum_{k=1}^\infty a_kB_k = \sum_{k=1}^\infty\left(\frac{\beta}{\alpha}\right)^k$$ and this sum, depending on our choice of $\alpha, \beta$, can either converge or diverge. For example, taking $\beta=2,\alpha=3$ means it converges, but taking $\beta=3,\alpha=2$ means it diverges.