# kinematic pairs

In a number of posts I swap the order of summation and resum. Since I tend to conceptualize the process of infinite series as a hot rodder would think about car engines in the 1950s, I think that a good metaphor would be of the kinematic pair, in particular where it states “A kinematic pair is the general name for two rigid bodies that can move with respect to each other via a mechanical constraint (joint) between the two bodies, with one or more degrees of freedom. “, so carrying on with the analogy:

“a kinematic pair is the name for a double series whose order of summation can move with respect to each other via a free variables constraint (joint) between the two series”

I’m going to make the (perhaps shameful) admission here that I don’t really care about convergence that much. The series I deal with tend to have ginormous denominators $n!^n!$, $n!n^{3n^7}$

Here’s an example of a kinematic pair:

$\sum_{m=2}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{m+n+1}}{n^{m}}$

>>> nsum(lambda m: nsum(lambda n: ((-1)**(m+n+1))/(n**m), [1, inf]), [2,inf])
mpf('0.386294361119890618834464242916358')


We can move one of those $-1$s outside, because it’s free in $n$:

$\sum_{m=2}^{\infty} (-1)^{m} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{m}}$

The inner is recognizably the Dirichlet eta function, so we can write:

$\sum_{m=2}^{\infty} (-1)^{m} \eta(m)$

>>> nsum(lambda m: ((-1)**m)*altzeta(m), [2,inf])
mpf('0.386294361119890618834464242916358')


But there’s an alternate way of looking at it, for if you take:

$\sum_{n=1}^{\infty} \sum_{m=2}^{\infty} \frac{(-1)^{m+n+1}}{n^{m}}$

And say “gee, that looks like a geometric series on the inside”, you can write:

$\sum_{n=1}^{\infty} (-1)^{n+1} \sum_{m=2}^{\infty} \frac{(-1)^{m}}{n^{m}}$

The inner sum is just $\frac{1}{n(n+1)}$, so we’ve got:

$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n(n+1)}$

>>> nsum(lambda n: ((-1)**(n+1))/(n*n+n), [1, inf])
mpf('0.386294361119890618834464242916358')


In the above, the joint/ mechanical constraint is $1/n^{m}$, because neither $n$ or $m$ is a free variable, so the two summations are free to ‘swing’ about it.

(Aside: according the inverse symbolic calculator,
$\sum_{n=1}^{\infty} \frac{1}{(n+1)2^{n}}$ is also a way of representing this number

>>> nsum(lambda m: 1/((m+1)*(2**m)), [1, inf])
mpf('0.386294361119890618834464242916358')


)

If I were faster on my toes, I would recognize that this constant is $\log(4)-1$