How to determine summability of "nested series" Consider some non-negative sequences $a_n, b_n, c_n,$ etc.
Suppose I have a "nested series" (not sure proper terminology)
$$\sum_{k = 1}^\infty a_k \sum_{j = 1}^k  b_j   \sum_{i = j}^k c_i$$
etc etc.
How should I go about determining the summability of this series?
Of course the first approach is to evaluate the inner term and work outwards...but is there a simpler way?
Suppose I know the summability of $a_k, b_k, c_k$ (e.g., summable, non summable, summable or non-summable, summable, summable). Is there a fast way of determining if the overall series is summable?
Any book that investigate these type of series would help :)
 A: 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.

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.

