# Proving $\pi \gt e+\frac{1}{e} \gt \pi-\frac{1}{\pi} \gt e$

I created this problem for myself as a fun exercise. I want to prove the following statement:

$$\pi \gt e+\dfrac{1}{e} \gt \pi-\dfrac{1}{\pi} \gt e$$

I found that the following upper/lower bounds for $e$ and $\pi$ are "good enough" to establish the above statement as true:

$$\dfrac{30}{11} \gt e \gt \dfrac{8}{3}$$

$$\dfrac{13}{4} \gt \pi \gt \dfrac{25}{8}$$

The upper/lower bounds for $e$ are easily proved by considering the series representation of $e^x$, and calculating partial sums for $x=1$ and $x=-1$.

However, I'm at a loss for how to establish the upper/lower bounds for $\pi$. I could approach it like Archimedes and use inscribed/circumscribed polygons (I believe it requires at least a $10$-gon and $18$-gon in this case). Is there an easier way to get these upper/lower bounds on $\pi$?

## EDIT:

I've also included the "alternative-proofs" tag because I am open to proofs of any kind, especially those which are elegant or particularly simple, and don't require knowledge of $e$ or $\pi$ to high precision.

• Continued fractions give you excellent rational approximations, by excess and by default. mathworld.wolfram.com/PiContinuedFraction.html – Yves Daoust Jun 16 '16 at 21:10
• My thought was that since $\Sigma_1^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$ and since all of the terms in this sum are positve, you could use it to find an approximation of $\pi$ – SquirtleSquad Jun 16 '16 at 21:13
• I don't think an upper bound of $3$ for $e$ is "good enough". $3+\frac 13 >\pi$ – Joffan Jun 16 '16 at 21:26
• What tools or methods are you constraining yourself to? Because if you can rattle off $e$ and $\pi$ to ten significant figures each, it's just a matter of arithmetic ... – John Jun 16 '16 at 21:50
• An upper bound of $11/4=2.75$ for $e$ would work along with the rest of your bounds. $14/5 = 2.8$ is too high. – Joffan Jun 16 '16 at 21:52

I will try to obtain the given inequalities with as few computations as I can. Our main tools will be the following series: \begin{align*} e^x&=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\\ \cosh(x):=\frac{e^x+e^{-x}}2&=1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\\ \arctan(x)&=x-\frac{x^3}3+\frac{x^5}5-+\cdots\quad(|x|<1) \end{align*}

We will also need the identity $\frac\pi4=\arctan(\frac12)+\arctan(\frac13)$. Proof: if $\alpha=\arctan(\frac12)$ and $\beta=\arctan(\frac13)$ then $$\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\frac{\frac12+\frac13}{1-\frac12\cdot\frac13}=1,$$ so $\alpha,\beta\in(0,\frac\pi4)$ implies $\alpha+\beta\in(0,\frac\pi2)$, so $\alpha+\beta=\arctan(1)=\frac\pi4$.

We can now proceed with the question.

Step 1: $\pi>e+\frac1e$. Now we have \begin{align*} e+\frac1e=2\cosh(1)&=2\left(1+\frac12+\frac1{2(3\cdot4)}+\frac1{2(3\cdot4)(5\cdot6)}+\cdots\right)\\ &<2\left(1+\frac12\left(1+\frac1{12}+\frac1{12^2}+\cdots\right)\right)\\ &=2\left(1+\frac12\cdot\frac{12}{11}\right)=\frac{34}{11}=3+\frac1{11}, \end{align*} and \begin{align*}\pi&=4(\arctan(\frac12)+\arctan(\frac13))\\ &\geq4\left(\frac12-\frac1{3\cdot2^3}+\frac13-\frac1{3\cdot3^3}\right)\\ &=\frac{505}{162}=3+\frac{19}{162}>3+\frac19. \end{align*} Combining the above gives the first inequality.

Step 2: $\pi-\frac1\pi>e$. This is immediate from the above, since $\pi>e+\frac1e>e$ implies $$\pi>e+\frac1e>e+\frac1\pi\implies\pi-\frac1\pi>e.$$

Step 3: $e+\frac1e>\pi-\frac1\pi$, or equivalently $\pi-e<\frac1e+\frac1\pi$. We have $$\pi=4(\arctan(\frac12)+\arctan(\frac13))<4(\frac12+\frac13)=\frac{10}3,$$ $$e>1+1+\frac1{2!}+\frac1{3!}+\frac1{4!}=\frac{65}{24},$$ $$\frac1e+\frac1\pi>\frac13+\frac1{\frac{10}3}=\frac{19}{30},$$ where $e<3$ follows from $e+\frac1e<3+\frac1{11}<3+\frac13$. Hence $$\pi-e<\frac{10}3-\frac{65}{24}=\frac58<\frac{19}{30}<\frac1e+\frac1\pi,$$ and we are done.

In practice, there are much more efficient ways of computing $\pi$ using the arctangent series, for instance Machin's formula $\frac\pi4=4\arctan(\frac15)-\arctan(\frac1{239})$, which already gives $\pi<4\cdot4\cdot\frac15=3.2$.

The approximations $3.14 < \pi < 3.34$ and $2.70 < e < 2.76$ suffice when doing the computation with two decimals.

I have a promising idea but still have to work the details.
There is a class of integrals that connect $\pi$ and $e$ that come from: $$\int_{0}^{+\infty}\frac{\cos(x)}{1+x^2}=\frac{\pi}{2e}\tag{1}$$

Now we may apply integration by parts multiple times, reaching: $$\frac{\pi}{e} = \int_{0}^{+\infty}\frac{p(x)(1-\cos x)}{(1+x^2)^k}\,dx \tag{2}$$ with $p(x)$ being some polynomial, then apply the Cauchy-Schwarz inequality to the RHS of $(2)$: $$\begin{eqnarray*} \frac{\pi^2}{e^2}&=&\left(\int_{0}^{+\infty}\frac{p(x)(1-\cos x)}{(1+x^2)^{k}}\,dx\right)^2\\&\leq& \left(\int_{0}^{+\infty}\frac{p(x)(1-\cos x)}{(1+x^2)^{k-1}}\,dx\right)\cdot\left(\int_{0}^{+\infty}\frac{p(x)(1-\cos x)}{(1+x^2)^{k+1}}\,dx\right)\tag{3}\end{eqnarray*}$$ leading to a product of two integrals that can still be evaluated in terms of $\pi$ and $e$.

That should give arbitrarily accurate approximations for the ratio $\frac{\pi}{e}$ and prove all the wanted inequalities. As I said, I will keep working on this approach. We also have: $$\int_{0}^{+\infty}\frac{\sin x}{x(1+x^2)}\,dx = \frac{\pi(e-1)}{2e}$$ where the integrand function in the LHS is waiting to be decomposed with respect to an orthogonal base of $L^2(\mathbb{R}^+)$ - I just have to understand with respect to which inner product.