Proving an inequality which leads to the proof of irrationality of $ \frac{1}{e}$

Question

I was trying to prove that $\ln\left(2\right)$ is irrational, but accidentally I ended up with the proof of irrationality of $\frac{1}{e}$.

Here is a proof of mine which has been inspired by the famous proof about irrationality of $e$ Which has been done by Joseph Fourier.

Proof:

It's known that :

$$1-\frac{1}{e}=\sum_{n=0}^{∞}\frac{\left(-1\right)^{n}}{\left(n+1\right)!}$$ using this it can be shown that:

$$\frac{1}{2}=\frac{\left(-1\right)^{0}}{1!}+\frac{\left(-1\right)^{1}}{2!}<\frac{\left(-1\right)^{0}}{1!}+\frac{\left(-1\right)^{1}}{2!}+...<\frac{\left(-1\right)^{0}}{\left(10\right)^{0}}+\frac{\left(-1\right)^{1}}{\left(10\right)^{1}}+...=\frac{10}{11}<1$$ Now assume $1-\frac{1}{e}$ is rational,in other words: $1-\frac{1}{e} =\frac{a}{b}$, where $a,b∈ℤ^+$ , also $b≠1$, because then $\frac{1}{2}<1-\frac{1}{e}=\frac{a}{b}=a<1$, which is a contradiction since $a∈ℤ^+$. Define:

$$x:=b!(1-\frac{1}{e}- \sum_{n=0}^{b}\frac{\left(-1\right)^{n}}{\left(n+1\right)!})$$

Plug the substitution $1-\frac{1}{e}=\frac{a}{b}$, $$x=b!( \frac{a}{b}- \sum_{n=0}^{b}\frac{\left(-1\right)^{n}}{\left(n+1\right)!})=a(b-1)!- \sum_{n=0}^{b}\frac{\left(-1\right)^{n}b!}{\left(n+1\right)!}$$

Since $\sum_{n=0}^{b}\frac{\left(-1\right)^{n}b!}{\left(n+1\right)!}∈ℤ$ and $ a(b-1)!∈ℤ$ implies $x∈ℤ$.

Rewrite $x$ such that:

$$x=b!( \sum_{n=0}^{∞}\frac{\left(-1\right)^{n}}{\left(n+1\right)!}- \sum_{n=0}^{b}\frac{\left(-1\right)^{n}}{\left(n+1\right)!})=\sum_{n=b+1}^{∞}\frac{\left(-1\right)^{n}b!}{\left(n+1\right)!}$$

Also $b$ is either odd or even, W.L.O.G let $b$ to be an odd number,then Since $b$ is odd then the lower limit of the sum starts with an even number,in other words:

$$\frac{b!}{(n+1)!}- \frac{b!}{(n+2)!}>0 ⇔(n+2)!>(n+1)!$$

Hence $$x=\sum_{n=b+1}^{∞}\frac{\left(-1\right)^{n}b!}{\left(n+1\right)!}>0$$

There are two cases to consider:

I)

where $n=2k$ for some $k∈ℤ$ with this assumption for all terms with $n ≥ b + 1$ we have the upper estimate:

$$\frac{b!}{\left(n+1\right)!}=\frac{1}{\left(b+1\right)\cdot\cdot\cdot\left(b+\left(n-b+1\right)\right)}<\frac{1}{\left(b+1\right)^{\left(n-b+1\right)}}$$

implies: $$\sum_{n=b+1}^{∞}\frac{b!}{\left(n+1\right)!}<\sum_{n=b+1}^{∞}\frac{1}{\left(b+1\right)^{\left(n-b+1\right)}}$$

Changing the index of summation to $n-b↦k$ we have the following relation:

$$\sum_{n=b+1}^{∞}\frac{b!}{\left(n+1\right)!}<\sum_{k=1}^{∞}\frac{1}{\left(b+1\right)^{\left(k+1\right)}}=\frac{1}{\left(b+1\right)^{2}}\left(\frac{1}{1-\frac{1}{b+1}}\right)=\frac{1}{b\left(b+1\right)}<\frac{1}{2}$$

now consider the following case:

II)

where $n=2k+1$ for some $k∈ℤ$ with this assumption for all terms with $n ≥ b + 2$ we have the upper estimate:

$$\frac{-\left(b!\right)}{\left(n+1\right)!}=\frac{-1}{\left(b+2\right)\cdot\cdot\cdot\left(b+\left(n-b+1\right)\right)}<0$$

implies:

$$\sum_{n=b+2}^{∞}\frac{-\left(b!\right)}{\left(n+1\right)!}<0$$

Using I and II , we conclude:

$$x=\sum\limits_{\substack{ {n=b+1} \\ \\ {n \enspace\text{even}} }}^\infty \frac{\left(b!\right)}{\left(n+1\right)!} +\sum\limits_{\substack{ {n=b+2} \\ \\ {n \enspace\text{odd}}}}^\infty \frac{-(b!)}{\left(n+1\right)!}<\frac{1}{2}$$ Finally it has been shown that $0<x<\frac{1}{2}$ , which is clearly a contradiction since $x∈ℤ$, implies

$$\boxed {1-\frac{1}{e}\ne\frac{a}{b}⇔1-\frac{1}{e}∈ℚ^{c}}\tag{1}$$ for all $b$ odd.

Now we have the same strategy for $b$ , when it's an even number:

Since $b$ is even then the lower limit of the sum starts with an odd number,in other words: $$\frac{-(b!)}{(n+1)!}+ \frac{b!}{(n+2)!}<0 ⇔(n+2)!>(n+1)!$$

Hence $$x=\sum_{n=b+1}^{∞}\frac{\left(-1\right)^{n}b!}{\left(n+1\right)!}<0$$

There are two cases to consider:

III)

where $n=2k+1$ for some $k∈ℤ$ with this assumption for all terms with $n ≥ b + 1$ we have the upper estimate:

$$\frac{-1}{\left(b+1\right)^{\left(n-b+1\right)}}<\frac{-1}{\left(b+1\right)\cdot\cdot\cdot\left(b+\left(n-b+1\right)\right)}=\frac{-(b!)}{\left(n+1\right)!}$$

implies: $$\sum_{n=b+1}^{∞}\frac{-1}{\left(b+1\right)^{\left(n-b+1\right)}}<\sum_{n=b+1}^{∞}\frac{-(b!)}{\left(n+1\right)!}$$

Changing the index of summation to $n-b↦k$ we can have the following relation:

$$\frac{-1}{2}<\frac{-1}{b\left(b+1\right)}=\frac{-1}{\left(b+1\right)^{2}}\left(\frac{1}{1-\frac{1}{b+1}}\right)=\sum_{k=1}^{∞}\frac{-1}{\left(b+1\right)^{\left(k+1\right)}}<\sum_{n=b+1}^{∞}\frac{-(b!)}{\left(n+1\right)!}$$

now consider the following case:

IV)

where $n=2k$ for some $k∈ℤ$ with this assumption for all terms with $n ≥ b + 2$ we have the upper estimate:

$$0<\frac{1}{\left(b+1\right)\cdot\cdot\cdot\left(b+\left(n-b+1\right)\right)}$$

implies:

$$0<\sum_{n=b+2}^{∞}\frac{\left(b!\right)}{\left(n+1\right)!}$$

Using III and IV , we conclude:

$$\frac{-1}{2}<x=\sum\limits_{\substack{ {n=b+1} \\ {n \enspace\text{odd}} }}^\infty \frac{-(b!)}{\left(n+1\right)!} +\sum\limits_{\substack{ {n=b+2} \\ {n \enspace\text{even}} }}^\infty \frac{b!}{\left(n+1\right)!}$$

Finally it has been shown that $\frac{-1}{2}<x<0$ , which is clearly a contradiction since $x∈ℤ$, implies

$$\boxed {1-\frac{1}{e}\ne\frac{a}{b}⇔1-\frac{1}{e}∈ℚ^{c}}\tag{2}$$ for all $b$ even.

From ${(1)}$ and ${(2)}$, it easily can be concluded that for neither $b$ odd nor $b$ even : $$\color{green} {\boxed {1-\frac{1}{e}\ne\frac{a}{b}⇔1-\frac{1}{e}∈ℚ^{c}}}$$

It's well-known that the sum of a rational and an irrational number is irrational , since $1$ is rational and $1-\frac{1}{e}$ is irrational , implies $-\frac{1}{e}$ is irratioal, the same can be said for $\frac{1}{e}. \qquad\blacksquare $

Also an upper and lower bound of $\frac{1}{e}$ can be concluded easily using the first inequality: $$0<\frac{\left(-1\right)^{2}}{2!}+...+\frac{\left(-1\right)^{\left(n+1\right)}}{\left(n+1\right)!}+...<\frac{1}{2} $$

The question is : How it can be shown that $$\frac{\left(-1\right)^{0}}{1!}+\frac{\left(-1\right)^{1}}{2!}+...+\frac{\left(-1\right)^{n}}{\left(n+1\right)!}+...<\frac{\left(-1\right)^{0}}{\left(10\right)^{0}}+\frac{\left(-1\right)^{1}}{\left(10\right)^{1}}+...=\frac{10}{11}<1$$

I know the inequality is true just because I know the value of $1-\frac{1}{e}$, but I want a proof which does not use this fact.

Updated: now I know my answer about the inequality but can someone determine whether my proof is right or it is not.

Answer 1

Your proof is not valid, but it can be easily fixed.

The following part has an error :

Define:$$x:=b!(1-\frac{1}{e}- \sum_{n=0}^{b}\frac{\left(-1\right)^{n}}{\left(n+1\right)!})$$Plug the substitution $1-\frac{1}{e}=\frac{a}{b}$, $$x=b!( \frac{a}{b}- \sum_{n=0}^{b}\frac{\left(-1\right)^{n}}{\left(n+1\right)!})=a(b-1)!- \sum_{n=0}^{b}\frac{\left(-1\right)^{n}b!}{\left(n+1\right)!}$$Since $\sum_{n=0}^{b}\frac{\left(-1\right)^{n}b!}{\left(n+1\right)!}∈ℤ$ and $ a(b-1)!∈ℤ$ implies $x∈ℤ$.

Note that $\sum_{n=0}^{b}\frac{\left(-1\right)^{n}b!}{\left(n+1\right)!}$ is not an integer since the last term of the sum is $\frac{(-1)^n}{b+1}$ which is not an integer.

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The following is a proof which is based on your idea.

Basically, all you need is to change the definition of $x$.

Let us define $x$ as follows :

$$x:=b!\bigg(1-\frac{1}{e}- \sum_{n=0}^{\color{red}{b-1}}\frac{\left(-1\right)^{n}}{\left(n+1\right)!}\bigg)$$Then, supposing that $1-\frac{1}{e}=\frac{a}{b}$, we get $$x=a(b-1)!- \sum_{n=0}^{b-1}\frac{\left(-1\right)^{n}b!}{\left(n+1\right)!}\quad (∈ℤ)$$

which can be written as

$$x=b!\bigg( \sum_{n=0}^{∞}\frac{\left(-1\right)^{n}}{\left(n+1\right)!}- \sum_{n=0}^{b-1}\frac{\left(-1\right)^{n}}{\left(n+1\right)!}\bigg)=\sum_{n=b}^{∞}\frac{\left(-1\right)^{n}b!}{\left(n+1\right)!}$$ $$=\sum_{\substack{ {n=b} \\ {n \enspace\text{even}} }}^{∞}\frac{b!}{\left(n+1\right)!}-\sum_{\substack{ {n=b} \\ {n \enspace\text{odd}} }}^{∞}\frac{b!}{\left(n+1\right)!}$$

Case 1 : If $b$ is odd, then $x$ is negative.

Since $$\frac{b!}{(n+1)!}=\frac{1}{(b+1)(b+2)\cdots (n+1)}\lt\frac{1}{(b+1)^{n-b+1}}\tag1$$ we get $$\sum_{\substack{ {n=b} \\ {n \enspace\text{odd}} }}^{∞}\frac{b!}{\left(n+1\right)!}<\sum_{m=1}^{∞}\frac{1}{\left(b+1\right)^{2m-1}}=\frac{b+1}{(b+1)^2-1}\lt\frac 12$$ We also have $$\sum_{\substack{ {n=b} \\ {n \enspace\text{even}} }}^{∞}\frac{b!}{\left(n+1\right)!}\gt 0$$ So, we get $$x=\sum_{\substack{ {n=b} \\ {n \enspace\text{even}} }}^{∞}\frac{b!}{\left(n+1\right)!}-\sum_{\substack{ {n=b} \\ {n \enspace\text{odd}} }}^{∞}\frac{b!}{\left(n+1\right)!}\gt -\frac 12$$ from which we have $-\frac 12\lt x\lt 0$, which contradicts that $x$ is an integer.

Case 2 : If $b$ is even, $x$ is positive.

From $(1)$, we get $$\sum_{\substack{ {n=b} \\ {n \enspace\text{even}} }}^{∞}\frac{b!}{\left(n+1\right)!}<\sum_{m=1}^{∞}\frac{1}{\left(b+1\right)^{2m-1}}=\frac{b+1}{(b+1)^2-1}\lt\frac 12$$ We also have $$\sum_{\substack{ {n=b} \\ {n \enspace\text{odd}} }}^{∞}\frac{b!}{\left(n+1\right)!}\gt 0$$ So, we get $$x=\sum_{\substack{ {n=b} \\ {n \enspace\text{even}} }}^{∞}\frac{b!}{\left(n+1\right)!}-\sum_{\substack{ {n=b} \\ {n \enspace\text{odd}} }}^{∞}\frac{b!}{\left(n+1\right)!}\lt \frac 12$$ from which we have $0\lt x\lt \frac 12$, which contradicts that $x$ is an integer.

Answer 2

Because $$\frac{\left(-1\right)^{0}}{1!}+\frac{\left(-1\right)^{1}}{2!}+...+\frac{\left(-1\right)^{n}}{\left(n+1\right)!}+... $$ is an alternating series, whose terms decrease in absolute value monotonically to zero,

it converges to a value less than that of the first term (which is $1$)

or the sum of the first three terms (which is $1-\frac12+\frac16=\frac23<\frac{10}{11}<1$).

Answer 3

Let me present my own proof of irrationality of $\,e\,$ or $\,e^{-1}\,$ or even $\,e^r\,$ for arbitrary rational $\,r\ne 0.\,$ I believe that my proof is original (let me know if you saw it in print or on the Internet).

For the above, it's enough to prove the irrationality of $\,e^{\frac {-1}q}\,$ for arbitrary positive integer $\,q.$

Given an arbitrary positive integer q, the arbitrary rational number $\,f\,$ is of the form

$$ f\,=\,f_{F\,n\,q}\,\,=\,\frac F{n!\cdot q^n} $$

where also $\,n\,$ is a positive integer, and $\,F\in\Bbb Z.$   Indeed, let $\, f:=\frac ab,\,$ where $\,a\,b\in\Bbb Z\,$ and $\,b>0.\ $ Then,

$$ f\,=\,f_{a\cdot(b-1)!\cdot q^b\,\, b\,\,q}\, =\,\,\frac{a\cdot(b-1)!\cdot q^b}{b!\cdot q^b} $$ Great!

On the other hand $$ e^{\frac{-1}q}\,=\,\sum_{n=0}^\infty \frac{(-1)^n}{n!\cdot q^n}\,=\,\, \lim_{n=\infty}\frac{E_n}{n!\cdot q^n} $$

where $\,E_n\,$ is the numerator of the $n$-th partial sum of the given series; and looking at the remainders of the above alternating series we know that $$ 0\, <\,\left|\frac{E_n}{n!\cdot q^n} - e^{\frac{-1}q}\right|\,<\,\frac 1{n!\cdot q^n} $$

In the light of above, $\,e^{\frac{-1}q}\,$ is not a rational number.   GREAT!

All you need to know was about the remainders of the exponential series -- the classical stuff. And nothing more, there was no additional computation.