Equation of the curve tangent to the unit circle and every circle with center $(\sum_{k=1}^{n-1}\frac1k, 0)$ and radius $\frac1n$, for $n>1$ 
I saw this problem on twitter where we have to find equation of red curve.
$C_{1}\implies x^2+y^2=1$
$C_{2}\implies (x-1)^2+y^2=\frac14$
$C_{3}\implies \Big(x-\frac32 \Big)^2+y^2=\frac19$
$C_{4}\implies \Big(x-\frac{11}{6} \Big)^2+y^2=\frac{1}{16}$
$\vdots$
$$C_{n}\implies \Bigg(x-\sum_{k=1}^{n-1}\frac{1}{k}\Bigg)^2+y^2=\frac{1}{n^2}$$ $\mathbf{\forall n>1}$
I thought from the shape of red curve that It should exponential function of form $\mathbf {y=a^{x-b}}$ , where $\mathbf{0<a<1}$.
But I don't know how to proceed further.
Thank you for your help!
 A: Not a full answer, but hopefully a helpful perspective:
We recall the surprisingly useful fact that the harmonic number $\sum_{k=1}^{n-1} \frac1k$ is equal to $p(n)$, where $p(z) = \frac{\Gamma'(z)}{\Gamma(z)} + \gamma$ is the logarithmic derivative of the Gamma function plus Euler's constant. This allows us to extend the discrete set of circles described on the OP to the continuous set of circles centered at $(p(z),0)$ with radius $\frac1z$ (for $z\ge1$). I'm guessing that the curve you seek is literally the upper envelope of this continuous family of curves.

The closest I could come to finding a formula for that upper envelope is to consider a fixed $y$-coordinate; the circle indexed by $z\ge1$ has equation $(x-p(z))^2+y^2=z^{-2}$, which means that the point furthest to the right at height $y$ is $\sqrt{z^{-2}-y^2}+p(z)$. Emperically, this function of $z$ (for fixed $y$) decreases to a unique maximum and then decreases (to $p(1/y)$ itself at $z=1/y$, where it stops being defined). So "all" we have to do is find the value $z = z(y)$ that maximizes $\sqrt{z^{-2}-y^2}+p(z)$; then the upper envelope would have equation $x=1/z(y)$.
