Question

If you had an infinite number of circles, each with different radii such that `r_n=1/n, ninZZ^+, nin[1,oo]`. What would the total area and circumference be of all the circles combined in terms of `pi`?

So, for area you would have `SigmaA=pi+pi/4+pi/9+pi/16+cdots+pi/oo^2` and for circumference you would have `SigmaC=2pi+pi+(2pi)/3+pi/2+(2pi)/5+cdots+(2pi)/oo`

Answer 1

Total area: `pi^3/6`

Total circumference: `oo`

Explanation:

In the following I use the fact that the harmonic sum to infinity diverges:

`sum_(n=1)^oo 1/n = oo`

and the sum of reciprocals of squares is `pi^2/6`

`sum_(n=1)^oo 1/n^2 = pi^2/6`

This second sum is not easy to prove. See https://socratic.org/s/aP6RvKGj

The circumference of a circle is `2pi r` where `r` is the radius.

So the total of all the circumferences of the circles of radii `1/n` would be:

`sum_(n=1)^oo 2pi(1/n) = 2pi sum_(n=1)^oo 1/n = oo`

(the sum diverges to `oo`)

The area of a circle is `pi r^2` where `r` is the radius.

So the total of all the areas of the circles of radii `1/n` would be:

`sum_(n=1)^oo pi(1/n)^2 = pi sum_(n=1)^oo 1/n^2 = pi * pi^2/6 = pi^3/6`