# A question about a proof of the irrationality of pi

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 $n, \dfrac{a^n\pi^{n+1}}{n!} \lt 1$

How did they get that "less than 1" bit?
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## Answers and Replies

fresh_42
Mentor
If ##\left(\dfrac{(a\pi)^n}{n!}\right)## tends to ##0##, what does that mean? For any ##\varepsilon > 0## there is ...

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fresh_42
Mentor
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.)

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?
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fresh_42
Mentor
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.

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}##?

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fresh_42
Mentor
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.

etotheipi
FactChecker
Science Advisor
Gold Member
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.