Proving that $sum_{n = 1}^{infty} frac{1}{n^s} = prod_{k=1}^{infty}frac{1}{1-p_k^{-s}}$
$begingroup$
Could you please have a look at my solution of the following exercise?
Show that for $s>1$ holds
$$sum_{n = 1}^{infty} frac{1}{n^s} = prod_{k=1}^{infty}frac{1}{1-p_k^{-s}}$$
where $(p_k)_k$ is the increasing sequence of primes.
I came up with the following solution:
At first we note that
$$prod_{k=1}^{infty}frac{1}{1-p_k^{-s}} = prod_{k=1}^{infty} frac{p_k^s}{p_k^s-1} = prod_{k=1}^{infty} biggl( sum_{j=0}^{infty} frac{1}{p_k^{s cdot j}}biggr)$$
Now we show both $ge$ and $le$:
$ge:$
$$prod_{k=1}^{N}frac{1}{1-p_k^{-s}} = prod_{k=1}^{N} frac{p_k^s}{p_k^s-1} $$
To make things easier we take care that the exponents are natural numbers
$$le prod_{k=1}^{N} frac{p_k^{lceil s rceil}}{p_k^{lfloor s rfloor} -1}$$
For fitting natural numbers $A,B$ we may write
$$ = frac{A}{B} = sum_{n=1}^{A} frac{1}{B} $$
And since $A ge B$
$$le sum_{n=1}^{A} frac{1}{n}$$
$le:$
We make the observation that in the result of the multiplication
$$biggl( sum_{i=0}^{a} frac{1}{p_1^i} biggr) cdot biggl( sum_{j=0}^{b} frac{1}{p_2^j}biggr) = sum_{i=0}^{a} sum_{j=0}^{b} frac{1}{p_1^i cdot p_2^j} $$
each combination of powers of $p_1$ and $p_2$ up to $p_1^a$ and $p_2^b$ appears exactly once in the denominators on the right side above. By generalising this and remembering about the fundametal theorem of arithmetic we obtain:
$$sum_{n = 1}^{N} frac{1}{n^s} le prod_{k=1}^{A} biggl( sum_{j=0}^{B_k} frac{1}{p_k^{s cdot j}}biggr) $$
for some natural numbers $A$ and $B$ and further:
$$le prod_{k=1}^{A} biggl( sum_{j=0}^{infty} frac{1}{p_k^{s cdot j}}biggr) = prod_{k=1}^{A} frac{p_k^s}{p_k^s-1} le prod_{k=1}^{infty} frac{p_k^s}{p_k^s-1}$$
Is this correct and if yes, could you tell me where exactly we need that $s>1$; I do not see why this is required.
sequences-and-series number-theory
asked Jan 6 at 16:47
3nondatur3nondatur
406112
$endgroup$
add a comment |
$begingroup$
Could you please have a look at my solution of the following exercise?
Show that for $s>1$ holds
$$sum_{n = 1}^{infty} frac{1}{n^s} = prod_{k=1}^{infty}frac{1}{1-p_k^{-s}}$$
where $(p_k)_k$ is the increasing sequence of primes.
I came up with the following solution:
At first we note that
$$prod_{k=1}^{infty}frac{1}{1-p_k^{-s}} = prod_{k=1}^{infty} frac{p_k^s}{p_k^s-1} = prod_{k=1}^{infty} biggl( sum_{j=0}^{infty} frac{1}{p_k^{s cdot j}}biggr)$$
Now we show both $ge$ and $le$:
$ge:$
$$prod_{k=1}^{N}frac{1}{1-p_k^{-s}} = prod_{k=1}^{N} frac{p_k^s}{p_k^s-1} $$
To make things easier we take care that the exponents are natural numbers
$$le prod_{k=1}^{N} frac{p_k^{lceil s rceil}}{p_k^{lfloor s rfloor} -1}$$
For fitting natural numbers $A,B$ we may write
$$ = frac{A}{B} = sum_{n=1}^{A} frac{1}{B} $$
And since $A ge B$
$$le sum_{n=1}^{A} frac{1}{n}$$
$le:$
We make the observation that in the result of the multiplication
$$biggl( sum_{i=0}^{a} frac{1}{p_1^i} biggr) cdot biggl( sum_{j=0}^{b} frac{1}{p_2^j}biggr) = sum_{i=0}^{a} sum_{j=0}^{b} frac{1}{p_1^i cdot p_2^j} $$
each combination of powers of $p_1$ and $p_2$ up to $p_1^a$ and $p_2^b$ appears exactly once in the denominators on the right side above. By generalising this and remembering about the fundametal theorem of arithmetic we obtain:
$$sum_{n = 1}^{N} frac{1}{n^s} le prod_{k=1}^{A} biggl( sum_{j=0}^{B_k} frac{1}{p_k^{s cdot j}}biggr) $$
for some natural numbers $A$ and $B$ and further:
$$le prod_{k=1}^{A} biggl( sum_{j=0}^{infty} frac{1}{p_k^{s cdot j}}biggr) = prod_{k=1}^{A} frac{p_k^s}{p_k^s-1} le prod_{k=1}^{infty} frac{p_k^s}{p_k^s-1}$$
Is this correct and if yes, could you tell me where exactly we need that $s>1$; I do not see why this is required.
sequences-and-series number-theory
asked Jan 6 at 16:47
3nondatur3nondatur
406112
$endgroup$
1
$begingroup$
Let $A_K$ be the set of integers whose prime factors are all $le p_K$. Then for $s > 0$, $prod_{k=1}^K frac{1}{1-p_k^{-s}} = prod_{k=1}^K (1+ sum_{j=1}^infty (p_k^j)^{-s}) = sum_{n in A_K} n^{-s}$. For $s > 1$ then $sum_{n=1}^infty n^{-s} - prod_{k=1}^K frac{1}{1-p_k^{-s}}=sum_{n not in A_K}n^{-s}le sum_{n=P_K+1}^infty n^{-s}$ whence $lim_{K to infty}sum_{n=1}^infty n^{-s}-prod_{k=1}^Kfrac{1}{1-p_k^{-s}} = 0$. For $sin (0,1]$ then $lim_{K to infty}prod_{k=1}^Kfrac{1}{1-p_k^{-s}}=lim_{K to infty}sum_{nin A_K}n^{-s}=sum_{n=1}^infty n^{-s}= infty$
$endgroup$
– reuns
Jan 6 at 17:36
2
$begingroup$
"could you tell me where exactly we need that $s>1$" Actually, we do not. When $sleqslant1$, the identity in the question holds as well, but it reads $infty=infty$.
$endgroup$
– Did
Jan 6 at 17:54
1
$begingroup$
See math.stackexchange.com/questions/427910/…?
$endgroup$
– Robert Z
Feb 14 at 11:45
add a comment |
$begingroup$
Could you please have a look at my solution of the following exercise?
Show that for $s>1$ holds
$$sum_{n = 1}^{infty} frac{1}{n^s} = prod_{k=1}^{infty}frac{1}{1-p_k^{-s}}$$
where $(p_k)_k$ is the increasing sequence of primes.
I came up with the following solution:
At first we note that
$$prod_{k=1}^{infty}frac{1}{1-p_k^{-s}} = prod_{k=1}^{infty} frac{p_k^s}{p_k^s-1} = prod_{k=1}^{infty} biggl( sum_{j=0}^{infty} frac{1}{p_k^{s cdot j}}biggr)$$
Now we show both $ge$ and $le$:
$ge:$
$$prod_{k=1}^{N}frac{1}{1-p_k^{-s}} = prod_{k=1}^{N} frac{p_k^s}{p_k^s-1} $$
To make things easier we take care that the exponents are natural numbers
$$le prod_{k=1}^{N} frac{p_k^{lceil s rceil}}{p_k^{lfloor s rfloor} -1}$$
For fitting natural numbers $A,B$ we may write
$$ = frac{A}{B} = sum_{n=1}^{A} frac{1}{B} $$
And since $A ge B$
$$le sum_{n=1}^{A} frac{1}{n}$$
$le:$
We make the observation that in the result of the multiplication
$$biggl( sum_{i=0}^{a} frac{1}{p_1^i} biggr) cdot biggl( sum_{j=0}^{b} frac{1}{p_2^j}biggr) = sum_{i=0}^{a} sum_{j=0}^{b} frac{1}{p_1^i cdot p_2^j} $$
each combination of powers of $p_1$ and $p_2$ up to $p_1^a$ and $p_2^b$ appears exactly once in the denominators on the right side above. By generalising this and remembering about the fundametal theorem of arithmetic we obtain:
$$sum_{n = 1}^{N} frac{1}{n^s} le prod_{k=1}^{A} biggl( sum_{j=0}^{B_k} frac{1}{p_k^{s cdot j}}biggr) $$
for some natural numbers $A$ and $B$ and further:
$$le prod_{k=1}^{A} biggl( sum_{j=0}^{infty} frac{1}{p_k^{s cdot j}}biggr) = prod_{k=1}^{A} frac{p_k^s}{p_k^s-1} le prod_{k=1}^{infty} frac{p_k^s}{p_k^s-1}$$
Is this correct and if yes, could you tell me where exactly we need that $s>1$; I do not see why this is required.
sequences-and-series number-theory
asked Jan 6 at 16:47
3nondatur3nondatur
406112
$endgroup$
Could you please have a look at my solution of the following exercise?
Show that for $s>1$ holds
$$sum_{n = 1}^{infty} frac{1}{n^s} = prod_{k=1}^{infty}frac{1}{1-p_k^{-s}}$$
where $(p_k)_k$ is the increasing sequence of primes.
I came up with the following solution:
At first we note that
$$prod_{k=1}^{infty}frac{1}{1-p_k^{-s}} = prod_{k=1}^{infty} frac{p_k^s}{p_k^s-1} = prod_{k=1}^{infty} biggl( sum_{j=0}^{infty} frac{1}{p_k^{s cdot j}}biggr)$$
Now we show both $ge$ and $le$:
$ge:$
$$prod_{k=1}^{N}frac{1}{1-p_k^{-s}} = prod_{k=1}^{N} frac{p_k^s}{p_k^s-1} $$
To make things easier we take care that the exponents are natural numbers
$$le prod_{k=1}^{N} frac{p_k^{lceil s rceil}}{p_k^{lfloor s rfloor} -1}$$
For fitting natural numbers $A,B$ we may write
$$ = frac{A}{B} = sum_{n=1}^{A} frac{1}{B} $$
And since $A ge B$
$$le sum_{n=1}^{A} frac{1}{n}$$
$le:$
We make the observation that in the result of the multiplication
$$biggl( sum_{i=0}^{a} frac{1}{p_1^i} biggr) cdot biggl( sum_{j=0}^{b} frac{1}{p_2^j}biggr) = sum_{i=0}^{a} sum_{j=0}^{b} frac{1}{p_1^i cdot p_2^j} $$
each combination of powers of $p_1$ and $p_2$ up to $p_1^a$ and $p_2^b$ appears exactly once in the denominators on the right side above. By generalising this and remembering about the fundametal theorem of arithmetic we obtain:
$$sum_{n = 1}^{N} frac{1}{n^s} le prod_{k=1}^{A} biggl( sum_{j=0}^{B_k} frac{1}{p_k^{s cdot j}}biggr) $$
for some natural numbers $A$ and $B$ and further:
$$le prod_{k=1}^{A} biggl( sum_{j=0}^{infty} frac{1}{p_k^{s cdot j}}biggr) = prod_{k=1}^{A} frac{p_k^s}{p_k^s-1} le prod_{k=1}^{infty} frac{p_k^s}{p_k^s-1}$$
Is this correct and if yes, could you tell me where exactly we need that $s>1$; I do not see why this is required.
sequences-and-series number-theory
sequences-and-series number-theory
asked Jan 6 at 16:47
3nondatur3nondatur
406112
asked Jan 6 at 16:47
3nondatur3nondatur
406112
edited Jan 30 at 15:38
3nondatur
asked Jan 6 at 16:47
3nondatur3nondatur
406112
asked Jan 6 at 16:47
3nondatur3nondatur
406112
asked Jan 6 at 16:47
3nondatur3nondatur
406112
406112
1
$begingroup$
Let $A_K$ be the set of integers whose prime factors are all $le p_K$. Then for $s > 0$, $prod_{k=1}^K frac{1}{1-p_k^{-s}} = prod_{k=1}^K (1+ sum_{j=1}^infty (p_k^j)^{-s}) = sum_{n in A_K} n^{-s}$. For $s > 1$ then $sum_{n=1}^infty n^{-s} - prod_{k=1}^K frac{1}{1-p_k^{-s}}=sum_{n not in A_K}n^{-s}le sum_{n=P_K+1}^infty n^{-s}$ whence $lim_{K to infty}sum_{n=1}^infty n^{-s}-prod_{k=1}^Kfrac{1}{1-p_k^{-s}} = 0$. For $sin (0,1]$ then $lim_{K to infty}prod_{k=1}^Kfrac{1}{1-p_k^{-s}}=lim_{K to infty}sum_{nin A_K}n^{-s}=sum_{n=1}^infty n^{-s}= infty$
$endgroup$
– reuns
Jan 6 at 17:36
2
$begingroup$
"could you tell me where exactly we need that $s>1$" Actually, we do not. When $sleqslant1$, the identity in the question holds as well, but it reads $infty=infty$.
$endgroup$
– Did
Jan 6 at 17:54
1
$begingroup$
See math.stackexchange.com/questions/427910/…?
$endgroup$
– Robert Z
Feb 14 at 11:45
add a comment |
1
$begingroup$
Let $A_K$ be the set of integers whose prime factors are all $le p_K$. Then for $s > 0$, $prod_{k=1}^K frac{1}{1-p_k^{-s}} = prod_{k=1}^K (1+ sum_{j=1}^infty (p_k^j)^{-s}) = sum_{n in A_K} n^{-s}$. For $s > 1$ then $sum_{n=1}^infty n^{-s} - prod_{k=1}^K frac{1}{1-p_k^{-s}}=sum_{n not in A_K}n^{-s}le sum_{n=P_K+1}^infty n^{-s}$ whence $lim_{K to infty}sum_{n=1}^infty n^{-s}-prod_{k=1}^Kfrac{1}{1-p_k^{-s}} = 0$. For $sin (0,1]$ then $lim_{K to infty}prod_{k=1}^Kfrac{1}{1-p_k^{-s}}=lim_{K to infty}sum_{nin A_K}n^{-s}=sum_{n=1}^infty n^{-s}= infty$
$endgroup$
– reuns
Jan 6 at 17:36
2
$begingroup$
"could you tell me where exactly we need that $s>1$" Actually, we do not. When $sleqslant1$, the identity in the question holds as well, but it reads $infty=infty$.
$endgroup$
– Did
Jan 6 at 17:54
1
$begingroup$
See math.stackexchange.com/questions/427910/…?
$endgroup$
– Robert Z
Feb 14 at 11:45
1
1
$begingroup$
Let $A_K$ be the set of integers whose prime factors are all $le p_K$. Then for $s > 0$, $prod_{k=1}^K frac{1}{1-p_k^{-s}} = prod_{k=1}^K (1+ sum_{j=1}^infty (p_k^j)^{-s}) = sum_{n in A_K} n^{-s}$. For $s > 1$ then $sum_{n=1}^infty n^{-s} - prod_{k=1}^K frac{1}{1-p_k^{-s}}=sum_{n not in A_K}n^{-s}le sum_{n=P_K+1}^infty n^{-s}$ whence $lim_{K to infty}sum_{n=1}^infty n^{-s}-prod_{k=1}^Kfrac{1}{1-p_k^{-s}} = 0$. For $sin (0,1]$ then $lim_{K to infty}prod_{k=1}^Kfrac{1}{1-p_k^{-s}}=lim_{K to infty}sum_{nin A_K}n^{-s}=sum_{n=1}^infty n^{-s}= infty$
$endgroup$
– reuns
Jan 6 at 17:36
$begingroup$
Let $A_K$ be the set of integers whose prime factors are all $le p_K$. Then for $s > 0$, $prod_{k=1}^K frac{1}{1-p_k^{-s}} = prod_{k=1}^K (1+ sum_{j=1}^infty (p_k^j)^{-s}) = sum_{n in A_K} n^{-s}$. For $s > 1$ then $sum_{n=1}^infty n^{-s} - prod_{k=1}^K frac{1}{1-p_k^{-s}}=sum_{n not in A_K}n^{-s}le sum_{n=P_K+1}^infty n^{-s}$ whence $lim_{K to infty}sum_{n=1}^infty n^{-s}-prod_{k=1}^Kfrac{1}{1-p_k^{-s}} = 0$. For $sin (0,1]$ then $lim_{K to infty}prod_{k=1}^Kfrac{1}{1-p_k^{-s}}=lim_{K to infty}sum_{nin A_K}n^{-s}=sum_{n=1}^infty n^{-s}= infty$
$endgroup$
– reuns
Jan 6 at 17:36
2
2
$begingroup$
"could you tell me where exactly we need that $s>1$" Actually, we do not. When $sleqslant1$, the identity in the question holds as well, but it reads $infty=infty$.
$endgroup$
– Did
Jan 6 at 17:54
$begingroup$
"could you tell me where exactly we need that $s>1$" Actually, we do not. When $sleqslant1$, the identity in the question holds as well, but it reads $infty=infty$.
$endgroup$
– Did
Jan 6 at 17:54
1
1
$begingroup$
See math.stackexchange.com/questions/427910/…?
$endgroup$
– Robert Z
Feb 14 at 11:45
$begingroup$
See math.stackexchange.com/questions/427910/…?
$endgroup$
– Robert Z
Feb 14 at 11:45
add a comment |
0
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$begingroup$
Let $A_K$ be the set of integers whose prime factors are all $le p_K$. Then for $s > 0$, $prod_{k=1}^K frac{1}{1-p_k^{-s}} = prod_{k=1}^K (1+ sum_{j=1}^infty (p_k^j)^{-s}) = sum_{n in A_K} n^{-s}$. For $s > 1$ then $sum_{n=1}^infty n^{-s} - prod_{k=1}^K frac{1}{1-p_k^{-s}}=sum_{n not in A_K}n^{-s}le sum_{n=P_K+1}^infty n^{-s}$ whence $lim_{K to infty}sum_{n=1}^infty n^{-s}-prod_{k=1}^Kfrac{1}{1-p_k^{-s}} = 0$. For $sin (0,1]$ then $lim_{K to infty}prod_{k=1}^Kfrac{1}{1-p_k^{-s}}=lim_{K to infty}sum_{nin A_K}n^{-s}=sum_{n=1}^infty n^{-s}= infty$
$endgroup$
– reuns
Jan 6 at 17:36
2
$begingroup$
"could you tell me where exactly we need that $s>1$" Actually, we do not. When $sleqslant1$, the identity in the question holds as well, but it reads $infty=infty$.
$endgroup$
– Did
Jan 6 at 17:54
1
$begingroup$
See math.stackexchange.com/questions/427910/…?
$endgroup$
– Robert Z
Feb 14 at 11:45