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?

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 $$\dot x(\theta)^2 + \dot y(\theta)^2 = (L - R \theta)^2.$$

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}