# How can one prove that $e<\pi$?

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 (${\tiny \text{which I personally find completely useless}}$).

My question will consist of 2 parts:

1. Prove (without calculator) that $e<\pi$;

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 $\pi$, otherwise the things would be very boring.

• Please do say how do you define both these numbers. – DonAntonio Sep 10 '13 at 16:18
• @DonAntonio Use any definition you like. – Start wearing purple Sep 10 '13 at 16:21
• I always thoght that comparing $e^\pi$ with $\pi^e$ (without calculator) was the only funny specimen of "this kind of problems". – Hagen von Eitzen Sep 10 '13 at 16:30
• @HagenvonEitzen somehow i never remember which is bigger and have to redo the argument everytime... – Jean-Sébastien Sep 10 '13 at 17:23
• @Jean-Sébastien Which grows faster, $x^e$ or $e^x$? Now, just remember that the inequality is the same for all $x$ (that is, $e^x$ is always greater than $x^e$). – Akiva Weinberger Dec 4 '15 at 20:01

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 $\pi$ as half the circumference of the unit circle.

Putting all this together we obtain $2\pi \gt 6$ or $\pi \gt 3$

We take $e$ as the sum $1+1+\frac 12+\frac 1{3!}+\cdots$ which converges absolutely and which, after the first three terms, is term by term less than the sum $1+1+\frac 12+\frac 1{2^2}+\cdots$ since the later terms in the second sum are obtained by dividing the previous term by $2$, and in the first sum by $n\gt 2$ (crudely for $n\ge 3$ we have $n!\gt 2^{n-1}$).

Summing the geometric series we have $e\lt 3 \lt\pi$.

What do we learn - well how easy it is to make an estimate depends on the definition. The geometric definition of $\pi$ 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 $\pi$ 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.

• Fast, simple, elegant... Love it! – Barranka Sep 17 '13 at 18:29
• Not to mention honest about the difficulties of using arc length in definitions of basic things. – Ryan Reich Sep 19 '13 at 23:26
• Note that it would be possible also to define $\frac {\pi}2$ as the least zero of the infinite series for $\cos x$ - which is, of course, an exponential series. And with some work it is possible to prove in this way that $\pi \gt 3$. However there is still some work to do to establish that the various different things called $\pi$ are actually the same. – Mark Bennet Sep 22 '13 at 20:00
• Very creative and interesting way to look at the problem. – Kirthi Raman Dec 24 '13 at 15:23

Use $e<3$ and $\pi>3$.

The first follows from $e:=\lim \left(1+\frac1n\right)^n$ quickly, the second from comparing a circle with its inscribed hexagon.

• can you give a hint to a quick proof of e<3 using the above definition? – Phani Raj Sep 10 '13 at 17:39
• Since $(1+1/n)^n<(1+1/n)^{n+1}$, it's enough to prove that the latter is decreasing. It starts from 4, then 3.375, then 3.16..., then 3.05..., then 2.98..., already smaller than 3. – pts Sep 10 '13 at 19:34
• addendum to @pts comment: to see that $$\left({{n+1}\over{n}}\right)^{{{1}\over{n+1}}}$$ is decreasing we show that $${{\left({{n+1}\over{n}}\right)^{{{1}\over{n+1}}}}\over{\left({{n+2 }\over{n+1}}\right)^{{{1}\over{n+2}}}}} \gt 1$$ or equivalently $$\left({{\left({{n+1}\over{n}}\right)^{{{1}\over{n+1}}}}\over{\left( {{n+2}\over{n+1}}\right)^{{{1}\over{n+2}}}}}\right)^{\left(n+1 \right)\,\left(n+2\right)} \gt 1$$ The latter is the product of two terms $\gt 1$ $$\left({n+1}\over{n} \right)$$ and $$\left({(n+1)^2}\over{n(n+2)} \right)^{(n+1)}$$ – miracle173 Sep 10 '13 at 22:51
• @miracle173: That is an impressive use of 600 characters – Eric Stucky Sep 22 '13 at 21:31

$$e =\sum_{n=0}^\infty \frac{1}{n!}= 2+\sum_{n=2}^\infty \frac{1}{n!}< 2+\sum_{n=2}^\infty \frac{1}{2^{n-1}}=3$$

By inscribing a regular hexagon in a circle, and noting its perimeter is less that that of the circle, we have $6r < 2 \pi r$ or $\pi > 3$.

Use absolutely convergent sum definitions for $e$ and $\pi$:

$$e^x=1+\sum_{n=1}^\infty{x^n \over n!}$$

and

$${\pi^2 \over 6} = \sum_{n=1}^\infty{1 \over n^2}$$

Then assume $e \ge \pi$, so $e^2 \ge \pi^2$, which means

$$1+\sum_{n=1}^\infty{2^n \over n!} \ge 6\sum_{n=1}^\infty{1 \over n^2}$$

$$\iff 1+\sum_{n=1}^\infty{{2^n\over n!}-{1\over n^2}} \ge 5\sum_{n=1}^\infty{1 \over n^2}$$

For all $n \ge 8$, ${2^n \over n!} \lt {1\over n^2}$, so $\sum_{n=8}^\infty{{2^n\over n!}-{1\over n^2}} \lt 0$. This means that

$$1+\sum_{n=1}^7{{2^n \over n!}-{1\over n^2}} \gt 5\sum_{n=1}^\infty{1\over n^2}$$

or more simply,

$$1+\sum_{n=1}^7{2^n \over n!} \gt 5\sum_{n=1}^\infty{1\over n^2}$$

$$\iff 1+2/1+4/2+8/6+16/24+32/120+64/720+128/5040 \gt 5+5/4+5/9+5/16+5/25+5/36+5/49+...$$

$$\iff 1/4+1/9+1/15+4/45+8/315 \gt 5/16+5/36+5/64+5/49+5/81+...$$

Which is clearly impossible given that paired LHS and RHS terms are all such that (LHS term) $\lt$ (RHS term), and that the LHS has only the terms shown. Therefore, our initial assumption is false, meaning that $e\lt\pi$.

It is difficult to say what can be learned from this. I think the clearest comparison would be if there were a definite limit statement for the value of $\pi$, rather than a geometric one, for which there is no "natural" geometric equivalent for $e$. Nevertheless, this sum comparison at least shows the difference between $\dfrac {2^n}{n!}$ and $\dfrac {1}{n^2}$.

Here is one direct way to demonstrate this inequality using only series:

$e=3 - \displaystyle\sum _{k=0}^{\infty}\dfrac{k+1}{(k+3)!}$

which is asymptotic from above, and

$\pi=3+2 \displaystyle\sum _{k=1}^{\infty } \frac{k (5 k+3) (2 k-1)! k!}{2^{k-1} (3 k+2)!}$

which is asymptotic from below.

In a glance, we see that $e$ equals 3 minus a positive quantity while $\pi$ equals 3 plus a positive quantity. Thus $e<3<\pi$ and $e<\pi$.

A benefit here is that the proof lends itself to being thought of in a dynamic sense; one can intuitively appreciate that as the series converge to their respective targets, they move away from the number 3 in opposite directions.

What is interesting about the question is that in mathematics, proving "obvious" things often requires an unexpected amount of thought. I am always struck by the wonderful range of approaches people employ in answering even simple questions. This underscores the fundamentally creative aspect of mathematics.

References for $e$ series:

References for $\pi$ series:

http://mathworld.wolfram.com/PiFormulas.html Eq (29) http://www.pi314.net/eng/schrogosper.php (States equivalence of nested expression and series) http://dspace.mit.edu/bitstream/handle/1721.1/6088/AIM-304.pdf?sequence=2

Another way of showing that $e<3$, using $\int_1^e \frac{1}{x} dx = 1$ (similar to the answer by dfeuer)

For any $x>0$ we have :

$$0\le (x-2)^2 = x^2-4x + 4 \implies \frac{1}{x} \ge 1-\frac{x}{4} \tag{1}$$

Then we can bound the integral by the area below the line: $$\int_1^3 \frac{1}{x} \, dx \ge \int_1^3 \left(1 - \frac{x}{4}\right) \, dx = 2 \frac{3/4 + 1/4}{2} =1 \tag{2}$$

This implies that $e \le 3$ (further, the inequality can be made strict by noting that $(1)$ is strict except for the single point $x=2$)

That $\pi >3$ can be seen easily by bounding the circle perimeter with an inscribed hexagon.

$\log e = 1$, where $$\log x = \int_1^x\!\frac 1 t\,dt.$$

By the hexagon approach (which I would frame in terms of estimating the integral $\int_{-1}^1\sqrt{1-x^2}\,dx$), $\pi > 3$, so $\log \pi > \log 3$.

Calculating the lower sum with an appropriate partition should do it.

Use the Leibniz formula for $\pi$: $$\frac{\pi}{4} = \sum_{j=0}^\infty \frac{(-1)^j}{2j+1} \text{.}$$ This is an alternating sum where each partial sum is alternately an upper and lower bound. The first few terms constrain $\pi$ to these ranges:

[0,4]
[2.666..., 4]
[2.666..., 3.4666...]
[2.895..., 3.4666...]
[2.895..., 3.33968...]
[2.9760..., 3.33968...]
[2.9760..., 3.28374...]
[3.01707..., 3.28374...]


at which point we know $\pi$ is greater than $3$.

A similar method is this descending series for $\mathrm{e}$: $$\mathrm{e} = 3 + \sum_{k=2}^\infty \frac{-1}{k!(k-1)k} \text{.}$$ Now after one term, we know $\mathrm{e} < 3-\frac{1}{4} < 3$ and the desired inequality follows.

For determining which of two unknown numbers given only by series is larger, it is helpful to have a series whose partial sums are bounds of some sort. Since we know we want $\mathrm{e} < \pi$ it is helpful to have a decreasing series for $\mathrm{e}$, an increasing series for $\pi$, and/or alternating series for either or both.

Further lesson: It is very nice to have intervals (or at least bounds) on unfamiliar numbers. Alternating series are quite handy for this.

Other series for this method: \begin{align} \mathrm{e}^{-1} &= \sum_{k=0}^\infty \frac{(-1)^k}{k!} &&\text{gives $\mathrm{e}<3$ after $5$ terms} \\ \frac{\pi^2}{8} &= \sum_{k=1}^\infty \frac{1}{(2k-1)^2} &&\text{gives $3 < \pi$ after 4 terms} \end{align}

Further further lesson: the more representations of a thing you know, the more you are able to show about it.
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