Euler's proof of divergence of sum of reciprocals of primes The 2019 Stack Overflow Developer Survey Results Are InSeries $sum fracsin(n)n cdot left(1+cdots +frac1nright)$ convergence questionConvergence or divergence of seriesSum of the recripocals of the Harmonic NumbersConvergence/Divergence of $sum_n=1^infty left(frac 1+cos(n)3 right)^n$can't determine the convergence/divergence hereInfinite sum including logarithmInfinite sum of each term of the harmonic series times a corresponding order p-seriesProve $frac12sum_p frac1p^2+frac13sum_p frac1p^3+cdots$ convergesDivergence of reciprocal of primes, EulerOn convergence of series of the generalized mean $sum_n=1^infty left(fraca_1^1/s+a_2^1/s+cdots +a_n^1/snright)^s.$
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Euler's proof of divergence of sum of reciprocals of primes
The 2019 Stack Overflow Developer Survey Results Are InSeries $sum fracsin(n)n cdot left(1+cdots +frac1nright)$ convergence questionConvergence or divergence of seriesSum of the recripocals of the Harmonic NumbersConvergence/Divergence of $sum_n=1^infty left(frac 1+cos(n)3 right)^n$can't determine the convergence/divergence hereInfinite sum including logarithmInfinite sum of each term of the harmonic series times a corresponding order p-seriesProve $frac12sum_p frac1p^2+frac13sum_p frac1p^3+cdots$ convergesDivergence of reciprocal of primes, EulerOn convergence of series of the generalized mean $sum_n=1^infty left(fraca_1^1/s+a_2^1/s+cdots +a_n^1/snright)^s.$
$begingroup$
On Wikipedia at link currently is:
beginalign
ln left( sum_n=1^infty frac1nright) & = lnleft( prod_p frac11-p^-1right)
= -sum_p ln left( 1-frac1pright) \
& = sum_p left( frac1p + frac12p^2 + frac13p^3 + cdots right) \
& = left( sum_pfrac1p right) + sum_p frac1p^2 left( frac12 + frac13p + frac14p^2 + cdots right) \
& < left( sum_p frac1p right) + sum_p frac1p^2 left( 1 + frac1p + frac1p^2 + cdots right) \
& = left( sum_p frac1p right) + left( sum_p frac1p(p-1) right) \
& = left( sum_p frac1p right) + C
endalign
and since $sum_n=1^infty frac1n$ diverges, so must $sum_p frac1p.$
Currently there is language like "audacious leaps of logic" and "correct result by questionable means".
Wouldn't this be a valid proof if we reversed it, though? If we assume $sum_p frac1p$ converges, can't we just go through the steps backwards and find $sum_n=1^infty frac1n$ converges, contradiction? Euler's work seems reasonable to me.
sequences-and-series prime-numbers fake-proofs
asked Jul 30 '14 at 3:29
user166998user166998
261
$endgroup$
add a comment |
$begingroup$
On Wikipedia at link currently is:
beginalign
ln left( sum_n=1^infty frac1nright) & = lnleft( prod_p frac11-p^-1right)
= -sum_p ln left( 1-frac1pright) \
& = sum_p left( frac1p + frac12p^2 + frac13p^3 + cdots right) \
& = left( sum_pfrac1p right) + sum_p frac1p^2 left( frac12 + frac13p + frac14p^2 + cdots right) \
& < left( sum_p frac1p right) + sum_p frac1p^2 left( 1 + frac1p + frac1p^2 + cdots right) \
& = left( sum_p frac1p right) + left( sum_p frac1p(p-1) right) \
& = left( sum_p frac1p right) + C
endalign
and since $sum_n=1^infty frac1n$ diverges, so must $sum_p frac1p.$
Currently there is language like "audacious leaps of logic" and "correct result by questionable means".
Wouldn't this be a valid proof if we reversed it, though? If we assume $sum_p frac1p$ converges, can't we just go through the steps backwards and find $sum_n=1^infty frac1n$ converges, contradiction? Euler's work seems reasonable to me.
sequences-and-series prime-numbers fake-proofs
asked Jul 30 '14 at 3:29
user166998user166998
261
$endgroup$
add a comment |
$begingroup$
On Wikipedia at link currently is:
beginalign
ln left( sum_n=1^infty frac1nright) & = lnleft( prod_p frac11-p^-1right)
= -sum_p ln left( 1-frac1pright) \
& = sum_p left( frac1p + frac12p^2 + frac13p^3 + cdots right) \
& = left( sum_pfrac1p right) + sum_p frac1p^2 left( frac12 + frac13p + frac14p^2 + cdots right) \
& < left( sum_p frac1p right) + sum_p frac1p^2 left( 1 + frac1p + frac1p^2 + cdots right) \
& = left( sum_p frac1p right) + left( sum_p frac1p(p-1) right) \
& = left( sum_p frac1p right) + C
endalign
and since $sum_n=1^infty frac1n$ diverges, so must $sum_p frac1p.$
Currently there is language like "audacious leaps of logic" and "correct result by questionable means".
Wouldn't this be a valid proof if we reversed it, though? If we assume $sum_p frac1p$ converges, can't we just go through the steps backwards and find $sum_n=1^infty frac1n$ converges, contradiction? Euler's work seems reasonable to me.
sequences-and-series prime-numbers fake-proofs
asked Jul 30 '14 at 3:29
user166998user166998
261
$endgroup$
On Wikipedia at link currently is:
beginalign
ln left( sum_n=1^infty frac1nright) & = lnleft( prod_p frac11-p^-1right)
= -sum_p ln left( 1-frac1pright) \
& = sum_p left( frac1p + frac12p^2 + frac13p^3 + cdots right) \
& = left( sum_pfrac1p right) + sum_p frac1p^2 left( frac12 + frac13p + frac14p^2 + cdots right) \
& < left( sum_p frac1p right) + sum_p frac1p^2 left( 1 + frac1p + frac1p^2 + cdots right) \
& = left( sum_p frac1p right) + left( sum_p frac1p(p-1) right) \
& = left( sum_p frac1p right) + C
endalign
and since $sum_n=1^infty frac1n$ diverges, so must $sum_p frac1p.$
Currently there is language like "audacious leaps of logic" and "correct result by questionable means".
Wouldn't this be a valid proof if we reversed it, though? If we assume $sum_p frac1p$ converges, can't we just go through the steps backwards and find $sum_n=1^infty frac1n$ converges, contradiction? Euler's work seems reasonable to me.
sequences-and-series prime-numbers fake-proofs
sequences-and-series prime-numbers fake-proofs
asked Jul 30 '14 at 3:29
user166998user166998
261
asked Jul 30 '14 at 3:29
user166998user166998
261
asked Jul 30 '14 at 3:29
user166998user166998
261
asked Jul 30 '14 at 3:29
user166998user166998
261
asked Jul 30 '14 at 3:29
user166998user166998
261
261
add a comment |
add a comment |
1 Answer
1
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$begingroup$
$
ln left( sum_n=1^infty frac1nright)$ has no meaning.
It is like you say $$ln (infty)=...$$
But $ln x$ is defined on real numbers and not on "anything we understand".
But everyone understands that Euler "understood" that the method was correct.
This is why the proof is attributed to him.
answered Nov 26 '15 at 18:37
Konstantinos GaitanasKonstantinos Gaitanas
6,79931938
$endgroup$
$begingroup$
You can assign $infty$ to diverging increasing sequences in which case the manipulations stated above are valid
$endgroup$
– reuns
Mar 30 at 10:36
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$
ln left( sum_n=1^infty frac1nright)$ has no meaning.
It is like you say $$ln (infty)=...$$
But $ln x$ is defined on real numbers and not on "anything we understand".
But everyone understands that Euler "understood" that the method was correct.
This is why the proof is attributed to him.
answered Nov 26 '15 at 18:37
Konstantinos GaitanasKonstantinos Gaitanas
6,79931938
$endgroup$
$begingroup$
You can assign $infty$ to diverging increasing sequences in which case the manipulations stated above are valid
$endgroup$
– reuns
Mar 30 at 10:36
add a comment |
$begingroup$
$
ln left( sum_n=1^infty frac1nright)$ has no meaning.
It is like you say $$ln (infty)=...$$
But $ln x$ is defined on real numbers and not on "anything we understand".
But everyone understands that Euler "understood" that the method was correct.
This is why the proof is attributed to him.
answered Nov 26 '15 at 18:37
Konstantinos GaitanasKonstantinos Gaitanas
6,79931938
$endgroup$
$begingroup$
You can assign $infty$ to diverging increasing sequences in which case the manipulations stated above are valid
$endgroup$
– reuns
Mar 30 at 10:36
add a comment |
$begingroup$
$
ln left( sum_n=1^infty frac1nright)$ has no meaning.
It is like you say $$ln (infty)=...$$
But $ln x$ is defined on real numbers and not on "anything we understand".
But everyone understands that Euler "understood" that the method was correct.
This is why the proof is attributed to him.
answered Nov 26 '15 at 18:37
Konstantinos GaitanasKonstantinos Gaitanas
6,79931938
$endgroup$
$
ln left( sum_n=1^infty frac1nright)$ has no meaning.
It is like you say $$ln (infty)=...$$
But $ln x$ is defined on real numbers and not on "anything we understand".
But everyone understands that Euler "understood" that the method was correct.
This is why the proof is attributed to him.
answered Nov 26 '15 at 18:37
Konstantinos GaitanasKonstantinos Gaitanas
6,79931938
answered Nov 26 '15 at 18:37
Konstantinos GaitanasKonstantinos Gaitanas
6,79931938
answered Nov 26 '15 at 18:37
Konstantinos GaitanasKonstantinos Gaitanas
6,79931938
answered Nov 26 '15 at 18:37
Konstantinos GaitanasKonstantinos Gaitanas
6,79931938
6,79931938
$begingroup$
You can assign $infty$ to diverging increasing sequences in which case the manipulations stated above are valid
$endgroup$
– reuns
Mar 30 at 10:36
add a comment |
$begingroup$
You can assign $infty$ to diverging increasing sequences in which case the manipulations stated above are valid
$endgroup$
– reuns
Mar 30 at 10:36
$begingroup$
You can assign $infty$ to diverging increasing sequences in which case the manipulations stated above are valid
$endgroup$
– reuns
Mar 30 at 10:36
$begingroup$
You can assign $infty$ to diverging increasing sequences in which case the manipulations stated above are valid
$endgroup$
– reuns
Mar 30 at 10:36
add a comment |
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