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#?

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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#

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Answer 1 · 2018-03-11T00:14:00

Total area: #pi^3/6#

Total circumference: #oo#

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#