Mixing
Evaluating $sum_{n=1}^infty frac{1}{(n^5)!} approx 1$ and proving that is irrational.
$begingroup$
Define $delta = sum_{n=1}^infty frac{1}{(n^5)!}$. Wolfram says it converges by the ratio test. Trying to prove that $delta$ is irrational, begin defining $S_n$ as:
begin{align}
S_n = (n^5)!delta : - (n^5)!sum_{k=1}^nfrac{1}{(k^5)!}
end{align}
Where $(n^5)!sum_{k=1}^nfrac{1}{(k^5)!}$ is an integer. Write $delta = 1/(1^5)!+1/(2^5)! + 1/(3^5)!+...+1/(n^5)!+1/(n+1)^5!+...$, so
begin{align}
S_n &= (n^5)!delta : - (n^5)!sum_{k=0}^nfrac{1}{(k^5)!}\
&=(n^5)!left(delta - sum_{k=0}^nfrac{1}{(k^5)!} right)\
&=(n^5)! left( frac{1}{(1^5)!}+frac{1}{(2^5)!}+...+frac{1}{(n^5)!}+
frac{1}{(n+1)^5!}+... - sum_{k=1}^nfrac{1}{(k^5)!} right)
end{align}
expanding the sum on the right it's possible to cancel a few terms:
begin{align}
S_n &= (n^5)! left( frac{1}{(1^5)!}+frac{1}{(2^5)!}+...+frac{1}{(n^5)!}+
frac{1}{(n+1)^5!}+... - frac{1}{(1^5)!} - frac{1}{(2^5)!}-...-frac{1}{(n^5)!}
right)\
&=(n^5)! left( frac{1}{(n+1)^5!}+frac{1}{(n+2)^5!}+... right)\
&= frac{(n^5)!}{(n+1)^5!}+frac{(n^5)!}{(n+2)^5!}+... \
end{align}
From this post we have that
begin{align}
S_n = frac{(n^5)!}{(n+1)^5!}+frac{(n^5)!}{(n+2)^5!}+cdots < frac{1}{n^5+1}
end{align}
So we have that $0<S_n<frac{1}{n^5+1}$. Assume that $delta = p/q$ then:
begin{align}
0< (n^5)!p : - q(n^5)!sum_{k=1}^nfrac{1}{(k^5)!} < frac{q}{n^5+1}
end{align}
So, for a large $n$, we found a integer between $0$ and $1$, meaning $delta$ is irrational
Wolfram says that $delta=1$, but then the proof above is wrong.
How to find $delta$ analytically? Numerically, also by wolf:
$delta approx sum_{n=1}^{10} 1/(n^5)! approx 1.000000000000000000000000000000000003800390754854743592594...$
proof-verification irrational-numbers
asked Jan 6 at 20:55
PintecoPinteco
731313
$endgroup$
add a comment |
$begingroup$
Define $delta = sum_{n=1}^infty frac{1}{(n^5)!}$. Wolfram says it converges by the ratio test. Trying to prove that $delta$ is irrational, begin defining $S_n$ as:
begin{align}
S_n = (n^5)!delta : - (n^5)!sum_{k=1}^nfrac{1}{(k^5)!}
end{align}
Where $(n^5)!sum_{k=1}^nfrac{1}{(k^5)!}$ is an integer. Write $delta = 1/(1^5)!+1/(2^5)! + 1/(3^5)!+...+1/(n^5)!+1/(n+1)^5!+...$, so
begin{align}
S_n &= (n^5)!delta : - (n^5)!sum_{k=0}^nfrac{1}{(k^5)!}\
&=(n^5)!left(delta - sum_{k=0}^nfrac{1}{(k^5)!} right)\
&=(n^5)! left( frac{1}{(1^5)!}+frac{1}{(2^5)!}+...+frac{1}{(n^5)!}+
frac{1}{(n+1)^5!}+... - sum_{k=1}^nfrac{1}{(k^5)!} right)
end{align}
expanding the sum on the right it's possible to cancel a few terms:
begin{align}
S_n &= (n^5)! left( frac{1}{(1^5)!}+frac{1}{(2^5)!}+...+frac{1}{(n^5)!}+
frac{1}{(n+1)^5!}+... - frac{1}{(1^5)!} - frac{1}{(2^5)!}-...-frac{1}{(n^5)!}
right)\
&=(n^5)! left( frac{1}{(n+1)^5!}+frac{1}{(n+2)^5!}+... right)\
&= frac{(n^5)!}{(n+1)^5!}+frac{(n^5)!}{(n+2)^5!}+... \
end{align}
From this post we have that
begin{align}
S_n = frac{(n^5)!}{(n+1)^5!}+frac{(n^5)!}{(n+2)^5!}+cdots < frac{1}{n^5+1}
end{align}
So we have that $0<S_n<frac{1}{n^5+1}$. Assume that $delta = p/q$ then:
begin{align}
0< (n^5)!p : - q(n^5)!sum_{k=1}^nfrac{1}{(k^5)!} < frac{q}{n^5+1}
end{align}
So, for a large $n$, we found a integer between $0$ and $1$, meaning $delta$ is irrational
Wolfram says that $delta=1$, but then the proof above is wrong.
How to find $delta$ analytically? Numerically, also by wolf:
$delta approx sum_{n=1}^{10} 1/(n^5)! approx 1.000000000000000000000000000000000003800390754854743592594...$
proof-verification irrational-numbers
asked Jan 6 at 20:55
PintecoPinteco
731313
$endgroup$
$begingroup$
Of course, $delta > 1$ because all the terms are positive and the first is $1$. Your proof that $delta$ is irrational looks fine to me, though I don't like using $n$ for both the dummy variable in the summation and the parameter for the $S_n$ terms.
$endgroup$
– Jason DeVito
Jan 6 at 21:12
$begingroup$
You surely don’t have $delta=1$ as the first term of the series is already equal to $1$.
$endgroup$
– mathcounterexamples.net
Jan 6 at 21:13
add a comment |
$begingroup$
Define $delta = sum_{n=1}^infty frac{1}{(n^5)!}$. Wolfram says it converges by the ratio test. Trying to prove that $delta$ is irrational, begin defining $S_n$ as:
begin{align}
S_n = (n^5)!delta : - (n^5)!sum_{k=1}^nfrac{1}{(k^5)!}
end{align}
Where $(n^5)!sum_{k=1}^nfrac{1}{(k^5)!}$ is an integer. Write $delta = 1/(1^5)!+1/(2^5)! + 1/(3^5)!+...+1/(n^5)!+1/(n+1)^5!+...$, so
begin{align}
S_n &= (n^5)!delta : - (n^5)!sum_{k=0}^nfrac{1}{(k^5)!}\
&=(n^5)!left(delta - sum_{k=0}^nfrac{1}{(k^5)!} right)\
&=(n^5)! left( frac{1}{(1^5)!}+frac{1}{(2^5)!}+...+frac{1}{(n^5)!}+
frac{1}{(n+1)^5!}+... - sum_{k=1}^nfrac{1}{(k^5)!} right)
end{align}
expanding the sum on the right it's possible to cancel a few terms:
begin{align}
S_n &= (n^5)! left( frac{1}{(1^5)!}+frac{1}{(2^5)!}+...+frac{1}{(n^5)!}+
frac{1}{(n+1)^5!}+... - frac{1}{(1^5)!} - frac{1}{(2^5)!}-...-frac{1}{(n^5)!}
right)\
&=(n^5)! left( frac{1}{(n+1)^5!}+frac{1}{(n+2)^5!}+... right)\
&= frac{(n^5)!}{(n+1)^5!}+frac{(n^5)!}{(n+2)^5!}+... \
end{align}
From this post we have that
begin{align}
S_n = frac{(n^5)!}{(n+1)^5!}+frac{(n^5)!}{(n+2)^5!}+cdots < frac{1}{n^5+1}
end{align}
So we have that $0<S_n<frac{1}{n^5+1}$. Assume that $delta = p/q$ then:
begin{align}
0< (n^5)!p : - q(n^5)!sum_{k=1}^nfrac{1}{(k^5)!} < frac{q}{n^5+1}
end{align}
So, for a large $n$, we found a integer between $0$ and $1$, meaning $delta$ is irrational
Wolfram says that $delta=1$, but then the proof above is wrong.
How to find $delta$ analytically? Numerically, also by wolf:
$delta approx sum_{n=1}^{10} 1/(n^5)! approx 1.000000000000000000000000000000000003800390754854743592594...$
proof-verification irrational-numbers
asked Jan 6 at 20:55
PintecoPinteco
731313
$endgroup$
Define $delta = sum_{n=1}^infty frac{1}{(n^5)!}$. Wolfram says it converges by the ratio test. Trying to prove that $delta$ is irrational, begin defining $S_n$ as:
begin{align}
S_n = (n^5)!delta : - (n^5)!sum_{k=1}^nfrac{1}{(k^5)!}
end{align}
Where $(n^5)!sum_{k=1}^nfrac{1}{(k^5)!}$ is an integer. Write $delta = 1/(1^5)!+1/(2^5)! + 1/(3^5)!+...+1/(n^5)!+1/(n+1)^5!+...$, so
begin{align}
S_n &= (n^5)!delta : - (n^5)!sum_{k=0}^nfrac{1}{(k^5)!}\
&=(n^5)!left(delta - sum_{k=0}^nfrac{1}{(k^5)!} right)\
&=(n^5)! left( frac{1}{(1^5)!}+frac{1}{(2^5)!}+...+frac{1}{(n^5)!}+
frac{1}{(n+1)^5!}+... - sum_{k=1}^nfrac{1}{(k^5)!} right)
end{align}
expanding the sum on the right it's possible to cancel a few terms:
begin{align}
S_n &= (n^5)! left( frac{1}{(1^5)!}+frac{1}{(2^5)!}+...+frac{1}{(n^5)!}+
frac{1}{(n+1)^5!}+... - frac{1}{(1^5)!} - frac{1}{(2^5)!}-...-frac{1}{(n^5)!}
right)\
&=(n^5)! left( frac{1}{(n+1)^5!}+frac{1}{(n+2)^5!}+... right)\
&= frac{(n^5)!}{(n+1)^5!}+frac{(n^5)!}{(n+2)^5!}+... \
end{align}
From this post we have that
begin{align}
S_n = frac{(n^5)!}{(n+1)^5!}+frac{(n^5)!}{(n+2)^5!}+cdots < frac{1}{n^5+1}
end{align}
So we have that $0<S_n<frac{1}{n^5+1}$. Assume that $delta = p/q$ then:
begin{align}
0< (n^5)!p : - q(n^5)!sum_{k=1}^nfrac{1}{(k^5)!} < frac{q}{n^5+1}
end{align}
So, for a large $n$, we found a integer between $0$ and $1$, meaning $delta$ is irrational
Wolfram says that $delta=1$, but then the proof above is wrong.
How to find $delta$ analytically? Numerically, also by wolf:
$delta approx sum_{n=1}^{10} 1/(n^5)! approx 1.000000000000000000000000000000000003800390754854743592594...$
proof-verification irrational-numbers
proof-verification irrational-numbers
asked Jan 6 at 20:55
PintecoPinteco
731313
asked Jan 6 at 20:55
PintecoPinteco
731313
asked Jan 6 at 20:55
PintecoPinteco
731313
asked Jan 6 at 20:55
PintecoPinteco
731313
asked Jan 6 at 20:55
PintecoPinteco
731313
731313
$begingroup$
Of course, $delta > 1$ because all the terms are positive and the first is $1$. Your proof that $delta$ is irrational looks fine to me, though I don't like using $n$ for both the dummy variable in the summation and the parameter for the $S_n$ terms.
$endgroup$
– Jason DeVito
Jan 6 at 21:12
$begingroup$
You surely don’t have $delta=1$ as the first term of the series is already equal to $1$.
$endgroup$
– mathcounterexamples.net
Jan 6 at 21:13
add a comment |
$begingroup$
Of course, $delta > 1$ because all the terms are positive and the first is $1$. Your proof that $delta$ is irrational looks fine to me, though I don't like using $n$ for both the dummy variable in the summation and the parameter for the $S_n$ terms.
$endgroup$
– Jason DeVito
Jan 6 at 21:12
$begingroup$
You surely don’t have $delta=1$ as the first term of the series is already equal to $1$.
$endgroup$
– mathcounterexamples.net
Jan 6 at 21:13
$begingroup$
Of course, $delta > 1$ because all the terms are positive and the first is $1$. Your proof that $delta$ is irrational looks fine to me, though I don't like using $n$ for both the dummy variable in the summation and the parameter for the $S_n$ terms.
$endgroup$
– Jason DeVito
Jan 6 at 21:12
$begingroup$
Of course, $delta > 1$ because all the terms are positive and the first is $1$. Your proof that $delta$ is irrational looks fine to me, though I don't like using $n$ for both the dummy variable in the summation and the parameter for the $S_n$ terms.
$endgroup$
– Jason DeVito
Jan 6 at 21:12
$begingroup$
You surely don’t have $delta=1$ as the first term of the series is already equal to $1$.
$endgroup$
– mathcounterexamples.net
Jan 6 at 21:13
$begingroup$
You surely don’t have $delta=1$ as the first term of the series is already equal to $1$.
$endgroup$
– mathcounterexamples.net
Jan 6 at 21:13
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A Liouville number is a real number x with the property that, for
every positive integer n, there exist integers p and q with q > 1 and
such that $$ 0<left|x-{frac {p}{q}}right|<{frac {1}{q^{n}}}. $$ A
Liouville number can thus be approximated "quite closely" by a
sequence of rational numbers. In 1844, Joseph Liouville showed that
all Liouville numbers are transcendental, thus establishing the
existence of transcendental numbers for the first time.
( https://en.wikipedia.org/wiki/Liouville_number)
Your number is clearly a Liouville number.
(I haven't checked your proof.)
answered Jan 6 at 21:28
Ethan BolkerEthan Bolker
42.4k549112
$endgroup$
add a comment |
Your Answer
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1 Answer
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1 Answer
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oldest
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$begingroup$
A Liouville number is a real number x with the property that, for
every positive integer n, there exist integers p and q with q > 1 and
such that $$ 0<left|x-{frac {p}{q}}right|<{frac {1}{q^{n}}}. $$ A
Liouville number can thus be approximated "quite closely" by a
sequence of rational numbers. In 1844, Joseph Liouville showed that
all Liouville numbers are transcendental, thus establishing the
existence of transcendental numbers for the first time.
( https://en.wikipedia.org/wiki/Liouville_number)
Your number is clearly a Liouville number.
(I haven't checked your proof.)
answered Jan 6 at 21:28
Ethan BolkerEthan Bolker
42.4k549112
$endgroup$
add a comment |
$begingroup$
A Liouville number is a real number x with the property that, for
every positive integer n, there exist integers p and q with q > 1 and
such that $$ 0<left|x-{frac {p}{q}}right|<{frac {1}{q^{n}}}. $$ A
Liouville number can thus be approximated "quite closely" by a
sequence of rational numbers. In 1844, Joseph Liouville showed that
all Liouville numbers are transcendental, thus establishing the
existence of transcendental numbers for the first time.
( https://en.wikipedia.org/wiki/Liouville_number)
Your number is clearly a Liouville number.
(I haven't checked your proof.)
answered Jan 6 at 21:28
Ethan BolkerEthan Bolker
42.4k549112
$endgroup$
add a comment |
$begingroup$
A Liouville number is a real number x with the property that, for
every positive integer n, there exist integers p and q with q > 1 and
such that $$ 0<left|x-{frac {p}{q}}right|<{frac {1}{q^{n}}}. $$ A
Liouville number can thus be approximated "quite closely" by a
sequence of rational numbers. In 1844, Joseph Liouville showed that
all Liouville numbers are transcendental, thus establishing the
existence of transcendental numbers for the first time.
( https://en.wikipedia.org/wiki/Liouville_number)
Your number is clearly a Liouville number.
(I haven't checked your proof.)
answered Jan 6 at 21:28
Ethan BolkerEthan Bolker
42.4k549112
$endgroup$
A Liouville number is a real number x with the property that, for
every positive integer n, there exist integers p and q with q > 1 and
such that $$ 0<left|x-{frac {p}{q}}right|<{frac {1}{q^{n}}}. $$ A
Liouville number can thus be approximated "quite closely" by a
sequence of rational numbers. In 1844, Joseph Liouville showed that
all Liouville numbers are transcendental, thus establishing the
existence of transcendental numbers for the first time.
( https://en.wikipedia.org/wiki/Liouville_number)
Your number is clearly a Liouville number.
(I haven't checked your proof.)
answered Jan 6 at 21:28
Ethan BolkerEthan Bolker
42.4k549112
answered Jan 6 at 21:28
Ethan BolkerEthan Bolker
42.4k549112
answered Jan 6 at 21:28
Ethan BolkerEthan Bolker
42.4k549112
answered Jan 6 at 21:28
Ethan BolkerEthan Bolker
42.4k549112
42.4k549112
add a comment |
add a comment |
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$begingroup$
Of course, $delta > 1$ because all the terms are positive and the first is $1$. Your proof that $delta$ is irrational looks fine to me, though I don't like using $n$ for both the dummy variable in the summation and the parameter for the $S_n$ terms.
$endgroup$
– Jason DeVito
Jan 6 at 21:12
$begingroup$
You surely don’t have $delta=1$ as the first term of the series is already equal to $1$.
$endgroup$
– mathcounterexamples.net
Jan 6 at 21:13