A question about a proof of the irrationality of pi

  • #1
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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:
[itex]\therefore \displaystyle{\lim_{n \to \infty}\left(\dfrac{a^n}{n!}\right)} = 0[/itex] and for sufficiently large [itex]
n, \dfrac{a^n\pi^{n+1}}{n!} \lt 1[/itex]


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

Answers and Replies

  • #2
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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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  • #3
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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.)
 
  • #4
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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?
.
 
  • #5
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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.
 
  • #6
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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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  • #7
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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.
 
  • #8
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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.
 

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