A very small disc winds around a larger circular disc of radius $R$ connected to it by a string. How long is the spiral it travels? A very small disc winds around a larger circular disc of radius $R$. It is connected to it by a string of length l that remains tight. What distance does it travel before it hits the larger disc of radius $R$ ? I've estimated that it is $2\pi(l-R)$ based on assuming a spiral is a little like a series of concentric circles.
 A: As stated on that physics page, it is easiest to consider the length and angle of the tight straight piece of rope.  If the angle $\theta$ starts at $0$ then winding the cord around gives a maximal angle $L/R$ (in radians).  The length of the straight piece of rope decreases linearly in $\theta$, so it is $$L(\theta) = L - R \theta.$$ Then the distance that the end of the rope travels is given by $$ \int_0^{L/R}L(\theta) \mathrm d \theta = \int_0^{L/R}(L - R \theta) \mathrm d \theta = \frac{L^2}{2R}.$$
This formula can also be obtained from an explicit parameterization of the path and calculating its length.  Again expressed in the angle $\theta$ the path can be given by $$\begin{eqnarray}
x(\theta) & = & R \sin(\theta) + (L - R \theta) \cos(\theta)\\
y(\theta) & = & -R \cos(\theta) + (L - R \theta) \sin(\theta)
\end{eqnarray}$$
The integrand of the path length integral simplifies quite a bit since $$\dot x(\theta)^2 + \dot y(\theta)^2 = (L - R \theta)^2.$$
A: Another way to add up your diminishing circle circumferences
is to use the formulas $2\pi R = \frac LN$ and  $N\pi = \frac{L}{2R},$
both of which are derived from $N = \frac{L}{2\pi R}.$
\begin{align}
\left(L - \frac LN\right)2\pi &+ \left(L - 2\frac LN\right)2\pi
 + \left(L - 3\frac LN\right)2\pi +  \cdots \\
&= \left(1 - \frac 1N\right)2\pi L + \left(1 - \frac 2N\right)2\pi L
 + \left(1 - \frac 3N\right)2\pi L +  \cdots \\
&= \left(N - 1\right)\frac{2\pi L}{N}
 + \left(N - 2\right)\frac{2\pi L}{N}
 + \left(N - 3\right)\frac{2\pi L}{N} +  \cdots \\
&= \left((N - 1) + (N - 2) + (N - 3) + \cdots + 1\right)\frac{2\pi L}{N} \\
&= \left(\frac{N(N - 1)}{2}\right)\frac{2\pi L}{N} \\
&= (N - 1)\pi L \\
&= N\pi L - \pi L \\
&= \frac{L^2}{2R} - \pi L. \\
\end{align}
But this is an underestimate, because at the beginning the end of the string is tracing a path of length approximately $L \theta$, where $\theta$ is the change in the angle of the string, although you have assumed it is only
$\left(1 - \frac 1N\right)L \theta.$
You could get a better approximation by wrapping the string around a regular polygon of $k$ sides and then letting $k$ increase to infinity
(effectively computing the integral of the path length as the string wraps around the circle);
the $\pi L$ term then drops out.

Your answer would also come up with the same sum except for a couple of errors.
The first error is that $2\pi \times 2\pi = 4\pi^2$ but you wrote $4\pi.$
The second error is on the very last step. Note that $2\pi N = \frac LR$
and $2\pi RN = L$ but $\pi(N + 1) = \frac{L}{2R} + \pi,$
so the result actually should be
\begin{align}
2\pi NL - 2\pi^2 RN(N+1)
 &= \frac{L^2}{R} - L\left(\frac{L}{2R} + \pi\right) \\
 &= \frac{L^2}{R} - \frac{L^2}{2R} - \pi L \\
 &= \frac{L^2}{2R} - \pi L.
\end{align}
A: This has been updated with corrections given in the accepted answer.
Assuming $L>>R$.
Path length approximated assuming that the spiral is approximately a sum of circles with circumference $2\pi R$ for each revolution of the spiral.
Assume $N=\frac{L}{2\pi R}$ is an integer in suitable units for the particular problem.
$\text{sum of diminishing circle circumferences}=$
$(L-2\pi R)2\pi + (L-2\pi R -2\pi R)2\pi +...$
$=2\pi NL - 4\pi^2 R (1+2 +3+...N)=2\pi NL - 4\pi^2 R(N(N+1))/2$
$=\frac{L^2}{R} - 4\pi^2 R\frac{(\frac{L^2}{(2\pi R)^2}+\frac{L}{2\pi R}))}{2}$
$=\frac{L^2}{R} - (\frac{L^2}{2R}+\pi L))$
$=\frac{L^2}{2R}-\pi L$
