A question about a proof of the irrationality of pi

[murshid_islam]

Here is the proof I was reading: https://mathschallenge.net/full/irrationality_of_pi

I have a question about this very last inequality at the end:

\therefore \displaystyle{\lim_{n \to \infty}\left(\dfrac{a^n}{n!}\right)} = 0 and for sufficiently large <br /> n, \dfrac{a^n\pi^{n+1}}{n!} \lt 1

How did they get that "less than 1" bit?
.

 

[fresh_42]

If $\left(\dfrac{(a\pi)^n}{n!}\right)$ tends to $0$, what does that mean? For any $\varepsilon > 0$ there is ...

 

[fresh_42]

You can use Stirling's formula and its lower boundary:
$$
\sqrt{2\pi n}\left(\dfrac{n^n}{e^n}\right) < n! \Longrightarrow\left(\dfrac{(a\pi)^n\cdot \pi}{n!}\right) < \left(\dfrac{a\pi e}{n}\right)^n\cdot \dfrac{\sqrt{\pi}}{\sqrt{2}}\cdot\dfrac{1}{\sqrt{n}}
$$
The first factor tends to $0$ and thus can be made smaller than $\dfrac{\sqrt{2}}{\sqrt{\pi}}.$

(If I made no mistake.)

 

[murshid_islam]

If $\left(\dfrac{(a\pi)^n}{n!}\right)$ tends to $0$, what does that mean? For any $\varepsilon > 0$ there is ...

I have been out of touch with that $\varepsilon-\delta$ stuff for about 15 years. Can you elaborate a little?
.

 

[fresh_42]

Formally: For any $\varepsilon > 0$ there is an $N(\varepsilon)$ - depending on that $\varepsilon$ - such that all $\left|\dfrac{(a\pi)^n}{n!}\right| < \varepsilon$ for all indices $n > N(\varepsilon)$.

This means that you can make $\dfrac{(a\pi)^n}{n!}$ arbitrary small, the greater $n$ is. So you can make it certainly smaller than $\dfrac{1}{\pi}$, if only the $n$ are large enough.

 

[murshid_islam]

Formally: For any $\varepsilon > 0$ there is an $N(\varepsilon)$ - depending on that $\varepsilon$ - such that all $\left|\dfrac{(a\pi)^n}{n!}\right| < \varepsilon$ for all indices $n > N(\varepsilon)$.

A general question about limits (just to check if I understood it):
if we wanted to find the limit of f(x) as $x \rightarrow -\infty$ (instead of $+\infty$), should we find $N(\varepsilon)$ such that |f(x) - the limit| $< \varepsilon$ for all $x < N(\varepsilon)$ instead of $x > N(\varepsilon)$.

This means that you can make $\dfrac{(a\pi)^n}{n!}$ arbitrary small, the greater $n$ is. So you can make it certainly smaller than $\dfrac{1}{\pi}$, if only the $n$ are large enough.

Does that mean that we can always find an $N(\varepsilon)$ such that any $n > N(\varepsilon)$ will lead to $\left|\frac{(a\pi)^n}{n!}\right| < \frac{1}{\pi}$?

 

[fresh_42]

A general question about limits (just to check if I understood it):
if we wanted to find the limit of f(x) as $x \rightarrow -\infty$ (instead of $+\infty$), should we find $N(\varepsilon)$ such that |f(x) - the limit| $< \varepsilon$ for all $x < N(\varepsilon)$ instead of $x > N(\varepsilon)$.

$\lim_{x \to a}f(x)=c$ means that for any sequence $(x_n)\longrightarrow a$ we have $\lim_{x_n \to \infty}f(x_n)=c.$ Here $a,c\in \mathbb{R}\cup \{\pm \infty\}.$ The $n$ are only a numbering of the sequence, i.e. it makes no sense to count them by negative numbers.

If $|c|<\infty$ is finite, then we have: For any sequence $(x_n)\longrightarrow a$ which converges to $a$, and any given, requested approximation $\varepsilon>0$, there is an $N(\varepsilon)$ such that $|f(x_n)-c|<\varepsilon$ for all $n>N(\varepsilon).$ Whether $a$ is infinitely positive, finite, or infinitely negative is irrelevant here; only that our (arbitrary) sequence $(x_n)$ converges to $a.$

If $c\in \{\pm \infty\}$, then we have again an arbitrary sequence $(x_n)\longrightarrow a$, for which the following holds: For any finite boundary $K$ there is a sequence index $N(K)$, such that $f(x_n) >K$ for all $n>N(K)$, in case $c=+\infty$, and $f(x_n) < K$ for all $n>N(K)$, in case $c=-\infty.$

If you forget about the function here, then the same definitions apply to $(x_n) \longrightarrow a.$

The indexing $n=1,2,3,\ldots$ is always positive. Sure, you may count by negative numbers, but that would only be unnecessarily confusing.

Continuity of the function $f$ and the $\varepsilon-\delta$ definition in comparison to the sequence definition of continuity is a different subject. Here we have only sequences.

 

[FactChecker]

How did they get that "less than 1" bit?

It is simpler than you are making it. Each increase of $n$ puts in another multiplier of $a*\pi/n$. Since $n$ can get huge, the product can get very small.