Is calculus 'ringing'?

The area of the ring is given by:

$\begin{aligned} A_R & = \left(\left(R+\frac 1R\right)^2 - R^2\right)\pi \\ & = \left(R^2+2+\frac 1{R^2} - R^2\right) \pi \\ & = \left(2+\frac 1{R^2} \right) \pi \end{aligned}$

We note that $A_R$ is minimum when $R \to \infty$ ; that is $A = \displaystyle \lim_{R \to \infty} \left(2+\frac 1{R^2} \right) \pi = 2\pi$ .

Now we have:

$\begin{aligned} S & = \sum_{n=1}^\infty \frac {(-1)^n}{(2n)!}\left(\frac A4 \right)^{2n} \\ & = \sum_{\color{#3D99F6}n=1}^\infty \frac {(-1)^n}{(2n)!}\left(\frac \pi 2 \right)^{2n} \\ & = \sum_{\color{#D61F06}n=0}^\infty \frac {(-1)^n}{(2n)!}\left(\frac \pi 2 \right)^{2n} \color{#D61F06}-1 & \small \color{#3D99F6} \text{Note that } \cos x = \sum_{n=0}^\infty \frac {(-1)^n}{(2n)!}x^{2n} \\ & = \cos \frac \pi 2 - 1 \\ & = 0 - 1 \\ & = \boxed{-1} \end{aligned}$