Question: This is Boas' exercise 1.15.31:

A tall tower of circular cross section is reinforced by horizontal circular disks (like large coins), one meter apart and of negligible thickness. The radius of the disk at height $n$ is $\frac{1}{n ln n}$ with $n ≥ 2$.

Here's the figure:

[![enter image description here](https://i.stack.imgur.com/8LLLr.png)](https://i.stack.imgur.com/8LLLr.png)

Assuming that the tower is of infinite height:

(a) Will the total area of the disks be finite or not? 

(b) If the disks are strengthened by wires going around their circumferences like tires, will the total length of wire required be finite or not?

Now for the first question comparing the area with the series $$\sum\_{n=0}^\infty \frac{1}{n^2}$$

We would understand that the series corresponding to the area in the question converges and therefore it will be finite.

For the second question we could easily use he integral test to see that the series corresponding to perimeter diverges to infinity.

And here's my question:

I understand that given the existence of $r^2$ in the formula for area, the area becomes smaller with a greater speed. But couldn't we assume the perimeter to be the rim of the disk and part of that area and ergo finite as well? What's wrong with my intuition?

Thank you in advance.

Answer: 
$$1 + {a^{-2^n}} = \frac{1 - a^{-2^{n+1}}}{1 - a^{-2^n}}$$
So we have a telescoping product
$$ \prod\_{n=0}^N \left(1 + a^{-2^n}\right) = \frac{1-a^{-2^{N+1}}}{1-a^{-1}}$$
An interesting way to look at this product is by realizing that it is tied to the binary representation of the integers spanning from $0$ to $2^{N+1}-1$. 

In fact, since $(1+x)(1+y)=1+x+y+xy$, then it is easy to deduct that $${\prod \_{n=0}^N \left(1+x^{2^n}\right)}={\sum\_{0 \le k \le 2^{N+1}-1}x^k}$$
i.e., where $k$ spans all the integers representable by $N$ bits.