How can one prove that e<πe<\pi?

This question is inspired by another one, asking to prove that something approximately equal to 1.2 is bigger than something approximately equal to 0.9. The numerical answer to this question was (expectedly) downvoted, though in my opinion it is the most reasonable approach to this kind of problems (which I personally find completely useless).

My question will consist of 2 parts:

  1. Prove (without calculator) that e<π;

  2. Explain what do we learn from the proof/what makes this problem interesting.

Edit: Existing answers only confirm my point of view about various weird inequalitites. Fortunately there is 3 between e and π, otherwise the things would be very boring.

Answer

Inscribe a regular hexagon in a circle of radius 1. Since a straight line is the shortest distance between two points the circumference of the circle is longer than the circumference of the hexagon. We take the definition of π as half the circumference of the unit circle.

Putting all this together we obtain 2π>6 or π>3

We take e as the sum 1+1+12+13!+ which converges absolutely and which, after the first three terms, is term by term less than the sum 1+1+12+122+ since the later terms in the second sum are obtained by dividing the previous term by 2, and in the first sum by n>2 (crudely for n3 we have n!>2n1).

Summing the geometric series we have e<3<π.

What do we learn - well how easy it is to make an estimate depends on the definition. The geometric definition of π lends itself to a good enough estimate. There are different ways of defining e too, but the sum offers a range of possibilities for estimating, particularly as the terms decrease very quickly. But the geometric definition for π requires assumed knowledge about a straight line as the shortest distance between two points, which seems obvious - yet conceals the trickiness of defining the length of a curve - so this looks simpler than it is.

Attribution
Source : Link , Question Author : Start wearing purple , Answer Author : Mark Bennet

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