I recently discovered that, if a chain of circles of radius $1/n^2$, where $n\in\mathbb{N}$, is tangent to the $x$-axis, then the the horizontal length of the chain is exactly $2$.
This can be shown by the fact that a circle of radius $1$ is tangent to the other side of the chain (which can be proven by using Descartes' Circle Theorem to show that if three circles of radius $1/n^2, 1$ and $\infty$ are mutually tangent, then a circle of radius $1/(n+1)^2$ is tangent to all three circles).
My question seeks to generalize this:
If a chain of circles of radius $1/n^p$, where $n\in\mathbb{N}$ and $p>2$ is an integer constant, is tangent to the $x$-axis, what is the horizontal length of the chain?
My attempt
Descartes' Circle Theorem seems to only apply when $p=2$. For other values of $p$, I don't know what curve is tangent to the other side of the chain.
Using the centres of two neighboring circles and Pythagorus' Theorem, the horizontal length of the chain is
$$\sum\limits_{k=1}^\infty 2(k^2+k)^{-p/2}$$
but I don't know how to evaluate this series.
To help you visualize, here is a chain of circles of radius $1/n^{3}$.
By using the formula near the bottom of this article we can decompose the summand for even $p=2n$ in partial fractions and explicitly evaluate the partial fraction coefficients for all $n$:
$$\frac{1}{k^n(k+1)^n}=\sum_{m=1}^n\frac{C_{m}}{k^m}+\sum_{m=1}^n\frac{D_{m}}{(k+1)^m}$$
$$C_m:=(-1)^m\frac{(n+m-1)!}{(n-m)!(n-1)!}~~~,~~~ D_m=(-1)^m C_m $$
which then allows us to write the requested sum as a finite sum over the values of zeta on the positive integers. Because of the special relationship between the $C,D$ coefficients, one can show explicitly that the values $\zeta(2\ell+1)$ do not appear in the final expressions. The sum takes, after some manipulations, the final form
$$\sum_{k=1}^\infty\frac{1}{k^n(k+1)^n}=\sum_{\ell=1}^n(-1)^{\ell+1}C_\ell+2\sum_{\ell=1}^{\lfloor n/2\rfloor}C_{2\ell}\zeta({2\ell})$$
Of course the zeta function at even integers has well known values in terms of the Bernoulli numbers, and in fact $\zeta(2n)\propto\pi^{2n}$. One can also derive general formulae for the family of sums $I_{m,n}=\sum_{k=1}^\infty\frac{1}{k^m(k+1)^n}$ in terms of $\zeta(k)$ in a similar manner as above, and also an explicit formula for their generating function.