summable family কী?

Example 7.10 (Summability by diagonal counting)

For which $s > 0$ is the family $\bigl((m + n)^{-s}\bigr)_{m, n \geq 1}$ summable? Group the finite partial sums by diagonals $m + n = k$: the diagonal $k$ carries $k - 1$ pairs, each contributing $k^{-s}$, so the finite sums are exactly bounded by (and exhaust)

a series with positive terms equivalent to $k^{1-s}$: summable iff $s - 1 > 1$, i.e. $s > 2$. The two-dimensional index eats one full power: a plane of terms is “one dimension more divergent” than a line — the counting geometry of the index set, not the size of individual terms, decides summability. (The same census shows $\bigl((m^2 + n^2)^{-1}\bigr)$ is not summable: on the diagonal $m + n = k$, each term is at least $k^{-2}$, and $(k-1)\cdot k^{-2}$ sums like the harmonic series.)

Example 7.12 (A rearrangement caught red-handed)

The alternating harmonic series sums to $\sum_{n\geq1}\frac{(-1)^{n-1}}{n} = \ln 2$ (Year 1 volume). Rearrange it as “one positive, two negatives”:

Grouping each block of three,

so the rearranged series converges to $\frac12\ln 2$ — half the original sum, with exactly the same terms. Non-absolutely convergent series remember the order of their terms; summable families are precisely the ones that do not.

Example 7.17 (A double-sum evaluation)

For $s > 1$ real, let $\zeta(s) = \sum_{n\geq1} n^{-s}$. Counting divisors by double summation — the family $(m^{-s}n^{-s})$ over $(m,n) \in (\N^*)^2$ is summable (product of convergent positive series) — and grouping by the product $q = mn$:

where $d(q)$ is the number of divisors of $q$. Summable families turn combinatorics into analysis.