\section*{Problem 2 - 161222}

A sphere \(K_{1}\) with a given radius \(r\) has a cylinder \(Z_{1}\) with a square cross-section inscribed in it.

This cylinder \(Z_{1}\) has a sphere \(K_{2}\) inscribed in it, and this sphere has a cylinder \(Z_{2}\) with a square cross-section inscribed in it.

This process is continued, i.e., if for a natural number \(n\) a sphere \(K_{n}\) and a cylinder \(Z_{n}\) with a square cross-section are already present, then a sphere \(K_{n+1}\) is inscribed in the cylinder \(Z_{n}\), and this sphere has a cylinder \(Z_{n+1}\) with a square cross-section inscribed in it.

For each \(n=1,2, \ldots\), let \(V_{n}\) be the volume of the sphere \(K_{n}\), and let \(S_{n}=V_{1}+V_{2}+\ldots+V_{n}\).

a) Determine the volume \(V_{10}\).

b) Determine \(S_{10}\).

c) Calculate \(\lim _{n \rightarrow \infty} S_{n}\), if this limit exists.

Hint:

A cylinder is said to be inscribed in a sphere if the circles bounding its two bases lie on the sphere. A sphere is said to be inscribed in a cylinder with a square cross-section if it touches the cylinder's two bases.
}

Let \(K\) be a sphere with radius \(R\), \(Z\) a cylinder inscribed in it with a square cross-section, and \(k\) a sphere inscribed in this cylinder with radius \(r\).

A planar cut that contains the axis of the cylinder produces a circular section with radius \(R\), in which a square is inscribed, and in which a circle with radius \(r\) is inscribed. Thus, \(2R\) is the length of the diagonal of the square, and \(2r\) is the side length of the square, so \(R = \sqrt{2} r\) or \(r = \frac{\sqrt{2}}{2} R\). If \(V = \frac{4}{3} \pi R^3\) is the volume of the sphere \(K\), then the volume \(v\) of \(k\) is exactly \(v = \frac{4}{3} \pi \cdot r^3 = V \cdot \frac{\sqrt{2}}{4}\).

Thus, for the problem,

\[
V_{n} = \frac{4}{3} \pi \cdot \left(\frac{\sqrt{2}}{4}\right)^{n-1} \cdot r^3
\]

and

\[
S_{n} = \frac{4}{3} \pi \cdot r^3 \cdot \left(\sum_{k=0}^{n-1} \left(\frac{\sqrt{2}}{4}\right)^k\right) = \frac{4}{3} \pi \cdot r^3 \cdot \frac{1 - \left(\frac{\sqrt{2}}{4}\right)^n}{1 - \left(\frac{\sqrt{2}}{4}\right)} = \frac{4}{3} \pi \cdot r^3 \cdot \frac{4^n - \sqrt{2}^n}{4^n - \sqrt{2} \cdot 4^{n-1}} = \frac{4}{3} \pi \cdot \frac{4^n - \sqrt{2}^n}{4^n - \sqrt{2} \cdot 4^{n-1}}
\]

In particular,

\[
V_{10} = \frac{4}{3} \pi \cdot \left(\frac{\sqrt{2}}{4}\right)^9 \cdot r^3 = \frac{2^2}{3} \cdot \pi \cdot \frac{2^4 \cdot \sqrt{2}}{2^{18}} \cdot r^3 = \frac{\sqrt{2}}{2^{12} \cdot 3} \cdot \pi \cdot r^3
\]

and

\[
S_{10} = \frac{4}{3} \pi \cdot \frac{4^{10} - \sqrt{2}}{4^{10} - \sqrt{2} \cdot 4^9} = \frac{2^2}{3} \pi \cdot \frac{2^{20} - 2^5}{2^{20} - \sqrt{2} \cdot 2^{18}} = \frac{1}{3} \pi \cdot \frac{2^{15} - 1}{2^{13} - \sqrt{2} \cdot 2^{11}} = \frac{1}{3} \pi \cdot \frac{32767}{8192 - 2048 \sqrt{2}}
\]

as well as

\[
\begin{aligned}
\lim_{n \to \infty} S_{n} = \lim_{n \to \infty} \frac{4}{3} \pi \cdot r^3 \cdot \frac{1 - \left(\frac{\sqrt{2}}{4}\right)^n}{1 - \left(\frac{\sqrt{2}}{4}\right)} & = \frac{4}{3} \pi \cdot r^3 \cdot \frac{1}{1 - \left(\frac{\sqrt{2}}{4}\right)} = \frac{4}{3} \pi \cdot r^3 \cdot \frac{4}{4 - \sqrt{2}} = \frac{16}{3} \pi \cdot r^3 \cdot \frac{4 + \sqrt{2}}{(4 - \sqrt{2})(4 + \sqrt{2})} = \\
& = \frac{16}{3} \pi \cdot r^3 \cdot \frac{4 + \sqrt{2}}{4^2 - 2^2} = \frac{16 + 4 \sqrt{2}}{9} \pi \cdot r^3
\end{aligned}
\]