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Help in understanding arithmetic form of Euler's proof of infinite primes.
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I recently saw this equation in a text book of mine and I am having a hard time following it. It shows how Euler's proof that the sum of the reciprocals of the primes diverge. This section deals with the fact that the original proof invokes real numbers and results from calculus, but then states that in principle a purely arithmetical proof can be given. It then proceeds to show that by starting with:
$$sum_i=0^∞ frac1p_i$$
Where the p's are primes.
Then says suppose this sums converges then we can choose an N so large that:
$$sum_i=N^∞ frac1p_i < 1$$
Let A be the set containing 1 together with those positive integers all of whose prime factors are among $p_1,...,p_N-1$. And let B be the set containing 1 together with those positive integers all of whose prime factors are among the $p_i$for $ile N$. We then have:
$$sum_n^∞ frac1n = sum_min A, k in B frac1mk =(sum_min A frac1m)(sum_k in B frac1k)$$
In this last product we can see that the first factor is bounded since it is a finite product from the above argument. We will then show that the second factor is bounded as well. From here the textbook loses me and I struggle to follow the logic. The textbook then says let $P(k_1,...,k_n,j_1,...,j_n,m)$ abbreviate the condition $k_1 + ... + k_n = m, k_1,...,k_n gt0, N le j_1 lt j_2 lt ... lt j_n$. Then:
$$sum_kin B frac1k = sum_m sum_P(k_1,...,k_n,j_1,...,j_n,m) frac1p^k_1_j_1...p^k_n_j_n= sum_m (sum_j=N^∞ frac1p_i)^m$$
The textbook then says:"The last sum (m summed over all natural numbers) converges by the condition on N. Thus we get that the harmonic series converges, and since this is not the case, it follows that $sum_i=0^∞ frac1p_i$ doesn't converge either". Though I understand the conclusion I struggle to follow how the sum of $frac1k$ is equal to those terms on the right hand side. Thanks for any help!.
sequences-and-series prime-numbers arithmetic proof-explanation
asked Jul 19 at 7:46
Brandon Lauwrens
162
add a comment |Â
up vote
1
down vote
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I recently saw this equation in a text book of mine and I am having a hard time following it. It shows how Euler's proof that the sum of the reciprocals of the primes diverge. This section deals with the fact that the original proof invokes real numbers and results from calculus, but then states that in principle a purely arithmetical proof can be given. It then proceeds to show that by starting with:
$$sum_i=0^∞ frac1p_i$$
Where the p's are primes.
Then says suppose this sums converges then we can choose an N so large that:
$$sum_i=N^∞ frac1p_i < 1$$
Let A be the set containing 1 together with those positive integers all of whose prime factors are among $p_1,...,p_N-1$. And let B be the set containing 1 together with those positive integers all of whose prime factors are among the $p_i$for $ile N$. We then have:
$$sum_n^∞ frac1n = sum_min A, k in B frac1mk =(sum_min A frac1m)(sum_k in B frac1k)$$
In this last product we can see that the first factor is bounded since it is a finite product from the above argument. We will then show that the second factor is bounded as well. From here the textbook loses me and I struggle to follow the logic. The textbook then says let $P(k_1,...,k_n,j_1,...,j_n,m)$ abbreviate the condition $k_1 + ... + k_n = m, k_1,...,k_n gt0, N le j_1 lt j_2 lt ... lt j_n$. Then:
$$sum_kin B frac1k = sum_m sum_P(k_1,...,k_n,j_1,...,j_n,m) frac1p^k_1_j_1...p^k_n_j_n= sum_m (sum_j=N^∞ frac1p_i)^m$$
The textbook then says:"The last sum (m summed over all natural numbers) converges by the condition on N. Thus we get that the harmonic series converges, and since this is not the case, it follows that $sum_i=0^∞ frac1p_i$ doesn't converge either". Though I understand the conclusion I struggle to follow how the sum of $frac1k$ is equal to those terms on the right hand side. Thanks for any help!.
sequences-and-series prime-numbers arithmetic proof-explanation
asked Jul 19 at 7:46
Brandon Lauwrens
162
The way you've defined $A$ and $B$, the only difference is that $i=N$ is included in $B$ but not in $A$. From the rest of the proof, it seems that it should say $ige N$ in the definition of $B$, not $ile N$. (Not sure whether that clears up your confusion or whether that was just an irrelevant typo.) Also, in the last displayed equation on the left-hand side, the indices $i$ and $j$ should be the same.
– joriki
Jul 19 at 11:13
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I recently saw this equation in a text book of mine and I am having a hard time following it. It shows how Euler's proof that the sum of the reciprocals of the primes diverge. This section deals with the fact that the original proof invokes real numbers and results from calculus, but then states that in principle a purely arithmetical proof can be given. It then proceeds to show that by starting with:
$$sum_i=0^∞ frac1p_i$$
Where the p's are primes.
Then says suppose this sums converges then we can choose an N so large that:
$$sum_i=N^∞ frac1p_i < 1$$
Let A be the set containing 1 together with those positive integers all of whose prime factors are among $p_1,...,p_N-1$. And let B be the set containing 1 together with those positive integers all of whose prime factors are among the $p_i$for $ile N$. We then have:
$$sum_n^∞ frac1n = sum_min A, k in B frac1mk =(sum_min A frac1m)(sum_k in B frac1k)$$
In this last product we can see that the first factor is bounded since it is a finite product from the above argument. We will then show that the second factor is bounded as well. From here the textbook loses me and I struggle to follow the logic. The textbook then says let $P(k_1,...,k_n,j_1,...,j_n,m)$ abbreviate the condition $k_1 + ... + k_n = m, k_1,...,k_n gt0, N le j_1 lt j_2 lt ... lt j_n$. Then:
$$sum_kin B frac1k = sum_m sum_P(k_1,...,k_n,j_1,...,j_n,m) frac1p^k_1_j_1...p^k_n_j_n= sum_m (sum_j=N^∞ frac1p_i)^m$$
The textbook then says:"The last sum (m summed over all natural numbers) converges by the condition on N. Thus we get that the harmonic series converges, and since this is not the case, it follows that $sum_i=0^∞ frac1p_i$ doesn't converge either". Though I understand the conclusion I struggle to follow how the sum of $frac1k$ is equal to those terms on the right hand side. Thanks for any help!.
sequences-and-series prime-numbers arithmetic proof-explanation
asked Jul 19 at 7:46
Brandon Lauwrens
162
I recently saw this equation in a text book of mine and I am having a hard time following it. It shows how Euler's proof that the sum of the reciprocals of the primes diverge. This section deals with the fact that the original proof invokes real numbers and results from calculus, but then states that in principle a purely arithmetical proof can be given. It then proceeds to show that by starting with:
$$sum_i=0^∞ frac1p_i$$
Where the p's are primes.
Then says suppose this sums converges then we can choose an N so large that:
$$sum_i=N^∞ frac1p_i < 1$$
Let A be the set containing 1 together with those positive integers all of whose prime factors are among $p_1,...,p_N-1$. And let B be the set containing 1 together with those positive integers all of whose prime factors are among the $p_i$for $ile N$. We then have:
$$sum_n^∞ frac1n = sum_min A, k in B frac1mk =(sum_min A frac1m)(sum_k in B frac1k)$$
In this last product we can see that the first factor is bounded since it is a finite product from the above argument. We will then show that the second factor is bounded as well. From here the textbook loses me and I struggle to follow the logic. The textbook then says let $P(k_1,...,k_n,j_1,...,j_n,m)$ abbreviate the condition $k_1 + ... + k_n = m, k_1,...,k_n gt0, N le j_1 lt j_2 lt ... lt j_n$. Then:
$$sum_kin B frac1k = sum_m sum_P(k_1,...,k_n,j_1,...,j_n,m) frac1p^k_1_j_1...p^k_n_j_n= sum_m (sum_j=N^∞ frac1p_i)^m$$
The textbook then says:"The last sum (m summed over all natural numbers) converges by the condition on N. Thus we get that the harmonic series converges, and since this is not the case, it follows that $sum_i=0^∞ frac1p_i$ doesn't converge either". Though I understand the conclusion I struggle to follow how the sum of $frac1k$ is equal to those terms on the right hand side. Thanks for any help!.
sequences-and-series prime-numbers arithmetic proof-explanation
asked Jul 19 at 7:46
Brandon Lauwrens
162
asked Jul 19 at 7:46
Brandon Lauwrens
162
asked Jul 19 at 7:46
Brandon Lauwrens
162
asked Jul 19 at 7:46
Brandon Lauwrens
162
162
The way you've defined $A$ and $B$, the only difference is that $i=N$ is included in $B$ but not in $A$. From the rest of the proof, it seems that it should say $ige N$ in the definition of $B$, not $ile N$. (Not sure whether that clears up your confusion or whether that was just an irrelevant typo.) Also, in the last displayed equation on the left-hand side, the indices $i$ and $j$ should be the same.
– joriki
Jul 19 at 11:13
add a comment |Â
The way you've defined $A$ and $B$, the only difference is that $i=N$ is included in $B$ but not in $A$. From the rest of the proof, it seems that it should say $ige N$ in the definition of $B$, not $ile N$. (Not sure whether that clears up your confusion or whether that was just an irrelevant typo.) Also, in the last displayed equation on the left-hand side, the indices $i$ and $j$ should be the same.
– joriki
Jul 19 at 11:13
The way you've defined $A$ and $B$, the only difference is that $i=N$ is included in $B$ but not in $A$. From the rest of the proof, it seems that it should say $ige N$ in the definition of $B$, not $ile N$. (Not sure whether that clears up your confusion or whether that was just an irrelevant typo.) Also, in the last displayed equation on the left-hand side, the indices $i$ and $j$ should be the same.
– joriki
Jul 19 at 11:13
The way you've defined $A$ and $B$, the only difference is that $i=N$ is included in $B$ but not in $A$. From the rest of the proof, it seems that it should say $ige N$ in the definition of $B$, not $ile N$. (Not sure whether that clears up your confusion or whether that was just an irrelevant typo.) Also, in the last displayed equation on the left-hand side, the indices $i$ and $j$ should be the same.
– joriki
Jul 19 at 11:13
add a comment |Â
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The way you've defined $A$ and $B$, the only difference is that $i=N$ is included in $B$ but not in $A$. From the rest of the proof, it seems that it should say $ige N$ in the definition of $B$, not $ile N$. (Not sure whether that clears up your confusion or whether that was just an irrelevant typo.) Also, in the last displayed equation on the left-hand side, the indices $i$ and $j$ should be the same.
– joriki
Jul 19 at 11:13