Solution to infinite product $prod_{p-primes}^{infty} frac{p}{p-1}$












1














I want to find the $prod_{p-primes}^{infty} frac{p}{p-1}$. This question stems from a question from amc 12a 2018. It goes as follows: Let $A$ be the set of positive integers that have no prime factors other than $2$, $3$, or $5$. The infinite sum
$$frac{1}{1} + frac{1}{2} + frac{1}{3} + frac{1}{4} + frac{1}{5} + frac{1}{6} + frac{1}{8} + frac{1}{9} + frac{1}{10} + frac{1}{12} + frac{1}{15} + frac{1}{16} + frac{1}{18} + frac{1}{20} + cdots$$
of the reciprocals of the elements of $A$ can be expressed as $frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$? Through some working out you can work out that the solution to this question is $sum_0^infty frac{1}{2^n}$$sum_0^infty frac{1}{3^n}$$sum_0^infty frac{1}{5^n}$, simple power series multiplication. But I started to wonder what the sum would be if $A$ was the set of positive integers that have no prime factors other than all primes. I think that the product will converge since $sum_1^infty log(a_n)$ seems to converge, but hey I could be wrong.










share|cite|improve this question


















  • 1




    en.wikipedia.org/wiki/Euler_product
    – R. Burton
    Nov 24 at 15:19
















1














I want to find the $prod_{p-primes}^{infty} frac{p}{p-1}$. This question stems from a question from amc 12a 2018. It goes as follows: Let $A$ be the set of positive integers that have no prime factors other than $2$, $3$, or $5$. The infinite sum
$$frac{1}{1} + frac{1}{2} + frac{1}{3} + frac{1}{4} + frac{1}{5} + frac{1}{6} + frac{1}{8} + frac{1}{9} + frac{1}{10} + frac{1}{12} + frac{1}{15} + frac{1}{16} + frac{1}{18} + frac{1}{20} + cdots$$
of the reciprocals of the elements of $A$ can be expressed as $frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$? Through some working out you can work out that the solution to this question is $sum_0^infty frac{1}{2^n}$$sum_0^infty frac{1}{3^n}$$sum_0^infty frac{1}{5^n}$, simple power series multiplication. But I started to wonder what the sum would be if $A$ was the set of positive integers that have no prime factors other than all primes. I think that the product will converge since $sum_1^infty log(a_n)$ seems to converge, but hey I could be wrong.










share|cite|improve this question


















  • 1




    en.wikipedia.org/wiki/Euler_product
    – R. Burton
    Nov 24 at 15:19














1












1








1


1





I want to find the $prod_{p-primes}^{infty} frac{p}{p-1}$. This question stems from a question from amc 12a 2018. It goes as follows: Let $A$ be the set of positive integers that have no prime factors other than $2$, $3$, or $5$. The infinite sum
$$frac{1}{1} + frac{1}{2} + frac{1}{3} + frac{1}{4} + frac{1}{5} + frac{1}{6} + frac{1}{8} + frac{1}{9} + frac{1}{10} + frac{1}{12} + frac{1}{15} + frac{1}{16} + frac{1}{18} + frac{1}{20} + cdots$$
of the reciprocals of the elements of $A$ can be expressed as $frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$? Through some working out you can work out that the solution to this question is $sum_0^infty frac{1}{2^n}$$sum_0^infty frac{1}{3^n}$$sum_0^infty frac{1}{5^n}$, simple power series multiplication. But I started to wonder what the sum would be if $A$ was the set of positive integers that have no prime factors other than all primes. I think that the product will converge since $sum_1^infty log(a_n)$ seems to converge, but hey I could be wrong.










share|cite|improve this question













I want to find the $prod_{p-primes}^{infty} frac{p}{p-1}$. This question stems from a question from amc 12a 2018. It goes as follows: Let $A$ be the set of positive integers that have no prime factors other than $2$, $3$, or $5$. The infinite sum
$$frac{1}{1} + frac{1}{2} + frac{1}{3} + frac{1}{4} + frac{1}{5} + frac{1}{6} + frac{1}{8} + frac{1}{9} + frac{1}{10} + frac{1}{12} + frac{1}{15} + frac{1}{16} + frac{1}{18} + frac{1}{20} + cdots$$
of the reciprocals of the elements of $A$ can be expressed as $frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$? Through some working out you can work out that the solution to this question is $sum_0^infty frac{1}{2^n}$$sum_0^infty frac{1}{3^n}$$sum_0^infty frac{1}{5^n}$, simple power series multiplication. But I started to wonder what the sum would be if $A$ was the set of positive integers that have no prime factors other than all primes. I think that the product will converge since $sum_1^infty log(a_n)$ seems to converge, but hey I could be wrong.







sequences-and-series infinite-product






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 24 at 15:07









Suchetan Dontha

11510




11510








  • 1




    en.wikipedia.org/wiki/Euler_product
    – R. Burton
    Nov 24 at 15:19














  • 1




    en.wikipedia.org/wiki/Euler_product
    – R. Burton
    Nov 24 at 15:19








1




1




en.wikipedia.org/wiki/Euler_product
– R. Burton
Nov 24 at 15:19




en.wikipedia.org/wiki/Euler_product
– R. Burton
Nov 24 at 15:19










3 Answers
3






active

oldest

votes


















2














You write "But I started to wonder what the sum would be if A
was the set of positive integers that have no prime factors other than all primes". Since any natural number can be written as a product of primes (unique up to reordering the factors) your set A will in this case be all natural numbers. So in this case your product can (in a formal sense) be rewritten as
$$ prod_{primes}frac{p}{p-1} = prod_{primes}frac{1}{1-1/p} = prod_{primes}sum_{i=0}^{infty} frac{1}{p^i} = sum_{n=1}^{infty} frac{1}{n} $$
To make sense of this formal manipulation, you could look at the function
$$ zeta(s) = sum_{n=1}^{infty} frac{1}{n^s} = prod_{primes}frac{p^s}{p^s-1} , (s>1)$$
which is known as the Riemann zeta function - and arguably the most famous function in all of mathematics. Your product would then correspond to the value of the Riemann zeta function as 1. As mentioned above the Riemann zeta-function has a pole at one (i.e. it is infinite), but one can could try to look at the asymptotic expansion of the sum
$$ sum_{n=1}^{N} frac{1}{n}$$
as $N$ goes to infinity. In this case something interesting does happen, namely
$$ lim_{N} (sum_{n=1}^{N}frac{1}{n} - ln(N)) = gamma$$
where $gamma$ is the Euler-Mascheroni constant. Now in your case one could argue that one should look not at the asymptotic expansion of $ sum_{n=1}^{N} frac{1}{n}$, but rather at the asymptotic expansion of
$$ prod_{p prime}^{p < N}frac{p}{p-1}$$
but I would expect the answer to be the same. In short - even though the product diverges doesn't mean something cool won't happen along the way.






share|cite|improve this answer





















  • Thank you for your response. I can't believe that I didn't recognize the product as the prime interpretation of the zeta function, thank you for pointing that out. Looking back now its pretty obvious that what I was asking for was the harmonic series. But I will look into the Euler-Mascheroni constant. I've seen it before and wondered how to get it.
    – Suchetan Dontha
    Nov 24 at 16:23



















3














${pover p-1}=1+{1over p-1}>1+frac1p$ and the sum of the reciprocals of the primes diverges, so the given product also diverges.






share|cite|improve this answer





















  • Thanks for your answer. I was hoping something cool would happen but I guess not.
    – Suchetan Dontha
    Nov 24 at 15:27



















0














An alternative approach: $frac{p}{p-1}geq 1+frac{1}{p}$ gives
$$ prod_{p}frac{p}{p-1}geq prod_{p}left(1+frac{1}{p}right) =!!!!!!!! sum_{substack{ngeq 1\ntext{ squarefree}}}!!!!!!!frac{1}{n} $$
but since the set of squarefree numbers has a positive density in $mathbb{N}$ ($frac{6}{pi^2}$, which is also the probability that two "random integers" are coprime), by summation by parts it follows that
$$sum_{substack{1leq nleq N\ntext{ squarefree}}}!!!!frac{1}{n}=O(1)+frac{6}{pi^2}sum_{n=1}^{N}frac{1}{n}=frac{6}{pi^2}log(N)+O(1) $$
so the original product is divergent.






share|cite|improve this answer





















    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3011661%2fsolution-to-infinite-product-prod-p-primes-infty-fracpp-1%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    You write "But I started to wonder what the sum would be if A
    was the set of positive integers that have no prime factors other than all primes". Since any natural number can be written as a product of primes (unique up to reordering the factors) your set A will in this case be all natural numbers. So in this case your product can (in a formal sense) be rewritten as
    $$ prod_{primes}frac{p}{p-1} = prod_{primes}frac{1}{1-1/p} = prod_{primes}sum_{i=0}^{infty} frac{1}{p^i} = sum_{n=1}^{infty} frac{1}{n} $$
    To make sense of this formal manipulation, you could look at the function
    $$ zeta(s) = sum_{n=1}^{infty} frac{1}{n^s} = prod_{primes}frac{p^s}{p^s-1} , (s>1)$$
    which is known as the Riemann zeta function - and arguably the most famous function in all of mathematics. Your product would then correspond to the value of the Riemann zeta function as 1. As mentioned above the Riemann zeta-function has a pole at one (i.e. it is infinite), but one can could try to look at the asymptotic expansion of the sum
    $$ sum_{n=1}^{N} frac{1}{n}$$
    as $N$ goes to infinity. In this case something interesting does happen, namely
    $$ lim_{N} (sum_{n=1}^{N}frac{1}{n} - ln(N)) = gamma$$
    where $gamma$ is the Euler-Mascheroni constant. Now in your case one could argue that one should look not at the asymptotic expansion of $ sum_{n=1}^{N} frac{1}{n}$, but rather at the asymptotic expansion of
    $$ prod_{p prime}^{p < N}frac{p}{p-1}$$
    but I would expect the answer to be the same. In short - even though the product diverges doesn't mean something cool won't happen along the way.






    share|cite|improve this answer





















    • Thank you for your response. I can't believe that I didn't recognize the product as the prime interpretation of the zeta function, thank you for pointing that out. Looking back now its pretty obvious that what I was asking for was the harmonic series. But I will look into the Euler-Mascheroni constant. I've seen it before and wondered how to get it.
      – Suchetan Dontha
      Nov 24 at 16:23
















    2














    You write "But I started to wonder what the sum would be if A
    was the set of positive integers that have no prime factors other than all primes". Since any natural number can be written as a product of primes (unique up to reordering the factors) your set A will in this case be all natural numbers. So in this case your product can (in a formal sense) be rewritten as
    $$ prod_{primes}frac{p}{p-1} = prod_{primes}frac{1}{1-1/p} = prod_{primes}sum_{i=0}^{infty} frac{1}{p^i} = sum_{n=1}^{infty} frac{1}{n} $$
    To make sense of this formal manipulation, you could look at the function
    $$ zeta(s) = sum_{n=1}^{infty} frac{1}{n^s} = prod_{primes}frac{p^s}{p^s-1} , (s>1)$$
    which is known as the Riemann zeta function - and arguably the most famous function in all of mathematics. Your product would then correspond to the value of the Riemann zeta function as 1. As mentioned above the Riemann zeta-function has a pole at one (i.e. it is infinite), but one can could try to look at the asymptotic expansion of the sum
    $$ sum_{n=1}^{N} frac{1}{n}$$
    as $N$ goes to infinity. In this case something interesting does happen, namely
    $$ lim_{N} (sum_{n=1}^{N}frac{1}{n} - ln(N)) = gamma$$
    where $gamma$ is the Euler-Mascheroni constant. Now in your case one could argue that one should look not at the asymptotic expansion of $ sum_{n=1}^{N} frac{1}{n}$, but rather at the asymptotic expansion of
    $$ prod_{p prime}^{p < N}frac{p}{p-1}$$
    but I would expect the answer to be the same. In short - even though the product diverges doesn't mean something cool won't happen along the way.






    share|cite|improve this answer





















    • Thank you for your response. I can't believe that I didn't recognize the product as the prime interpretation of the zeta function, thank you for pointing that out. Looking back now its pretty obvious that what I was asking for was the harmonic series. But I will look into the Euler-Mascheroni constant. I've seen it before and wondered how to get it.
      – Suchetan Dontha
      Nov 24 at 16:23














    2












    2








    2






    You write "But I started to wonder what the sum would be if A
    was the set of positive integers that have no prime factors other than all primes". Since any natural number can be written as a product of primes (unique up to reordering the factors) your set A will in this case be all natural numbers. So in this case your product can (in a formal sense) be rewritten as
    $$ prod_{primes}frac{p}{p-1} = prod_{primes}frac{1}{1-1/p} = prod_{primes}sum_{i=0}^{infty} frac{1}{p^i} = sum_{n=1}^{infty} frac{1}{n} $$
    To make sense of this formal manipulation, you could look at the function
    $$ zeta(s) = sum_{n=1}^{infty} frac{1}{n^s} = prod_{primes}frac{p^s}{p^s-1} , (s>1)$$
    which is known as the Riemann zeta function - and arguably the most famous function in all of mathematics. Your product would then correspond to the value of the Riemann zeta function as 1. As mentioned above the Riemann zeta-function has a pole at one (i.e. it is infinite), but one can could try to look at the asymptotic expansion of the sum
    $$ sum_{n=1}^{N} frac{1}{n}$$
    as $N$ goes to infinity. In this case something interesting does happen, namely
    $$ lim_{N} (sum_{n=1}^{N}frac{1}{n} - ln(N)) = gamma$$
    where $gamma$ is the Euler-Mascheroni constant. Now in your case one could argue that one should look not at the asymptotic expansion of $ sum_{n=1}^{N} frac{1}{n}$, but rather at the asymptotic expansion of
    $$ prod_{p prime}^{p < N}frac{p}{p-1}$$
    but I would expect the answer to be the same. In short - even though the product diverges doesn't mean something cool won't happen along the way.






    share|cite|improve this answer












    You write "But I started to wonder what the sum would be if A
    was the set of positive integers that have no prime factors other than all primes". Since any natural number can be written as a product of primes (unique up to reordering the factors) your set A will in this case be all natural numbers. So in this case your product can (in a formal sense) be rewritten as
    $$ prod_{primes}frac{p}{p-1} = prod_{primes}frac{1}{1-1/p} = prod_{primes}sum_{i=0}^{infty} frac{1}{p^i} = sum_{n=1}^{infty} frac{1}{n} $$
    To make sense of this formal manipulation, you could look at the function
    $$ zeta(s) = sum_{n=1}^{infty} frac{1}{n^s} = prod_{primes}frac{p^s}{p^s-1} , (s>1)$$
    which is known as the Riemann zeta function - and arguably the most famous function in all of mathematics. Your product would then correspond to the value of the Riemann zeta function as 1. As mentioned above the Riemann zeta-function has a pole at one (i.e. it is infinite), but one can could try to look at the asymptotic expansion of the sum
    $$ sum_{n=1}^{N} frac{1}{n}$$
    as $N$ goes to infinity. In this case something interesting does happen, namely
    $$ lim_{N} (sum_{n=1}^{N}frac{1}{n} - ln(N)) = gamma$$
    where $gamma$ is the Euler-Mascheroni constant. Now in your case one could argue that one should look not at the asymptotic expansion of $ sum_{n=1}^{N} frac{1}{n}$, but rather at the asymptotic expansion of
    $$ prod_{p prime}^{p < N}frac{p}{p-1}$$
    but I would expect the answer to be the same. In short - even though the product diverges doesn't mean something cool won't happen along the way.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 24 at 16:12









    Testcase

    1362




    1362












    • Thank you for your response. I can't believe that I didn't recognize the product as the prime interpretation of the zeta function, thank you for pointing that out. Looking back now its pretty obvious that what I was asking for was the harmonic series. But I will look into the Euler-Mascheroni constant. I've seen it before and wondered how to get it.
      – Suchetan Dontha
      Nov 24 at 16:23


















    • Thank you for your response. I can't believe that I didn't recognize the product as the prime interpretation of the zeta function, thank you for pointing that out. Looking back now its pretty obvious that what I was asking for was the harmonic series. But I will look into the Euler-Mascheroni constant. I've seen it before and wondered how to get it.
      – Suchetan Dontha
      Nov 24 at 16:23
















    Thank you for your response. I can't believe that I didn't recognize the product as the prime interpretation of the zeta function, thank you for pointing that out. Looking back now its pretty obvious that what I was asking for was the harmonic series. But I will look into the Euler-Mascheroni constant. I've seen it before and wondered how to get it.
    – Suchetan Dontha
    Nov 24 at 16:23




    Thank you for your response. I can't believe that I didn't recognize the product as the prime interpretation of the zeta function, thank you for pointing that out. Looking back now its pretty obvious that what I was asking for was the harmonic series. But I will look into the Euler-Mascheroni constant. I've seen it before and wondered how to get it.
    – Suchetan Dontha
    Nov 24 at 16:23











    3














    ${pover p-1}=1+{1over p-1}>1+frac1p$ and the sum of the reciprocals of the primes diverges, so the given product also diverges.






    share|cite|improve this answer





















    • Thanks for your answer. I was hoping something cool would happen but I guess not.
      – Suchetan Dontha
      Nov 24 at 15:27
















    3














    ${pover p-1}=1+{1over p-1}>1+frac1p$ and the sum of the reciprocals of the primes diverges, so the given product also diverges.






    share|cite|improve this answer





















    • Thanks for your answer. I was hoping something cool would happen but I guess not.
      – Suchetan Dontha
      Nov 24 at 15:27














    3












    3








    3






    ${pover p-1}=1+{1over p-1}>1+frac1p$ and the sum of the reciprocals of the primes diverges, so the given product also diverges.






    share|cite|improve this answer












    ${pover p-1}=1+{1over p-1}>1+frac1p$ and the sum of the reciprocals of the primes diverges, so the given product also diverges.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 24 at 15:24









    saulspatz

    13.7k21328




    13.7k21328












    • Thanks for your answer. I was hoping something cool would happen but I guess not.
      – Suchetan Dontha
      Nov 24 at 15:27


















    • Thanks for your answer. I was hoping something cool would happen but I guess not.
      – Suchetan Dontha
      Nov 24 at 15:27
















    Thanks for your answer. I was hoping something cool would happen but I guess not.
    – Suchetan Dontha
    Nov 24 at 15:27




    Thanks for your answer. I was hoping something cool would happen but I guess not.
    – Suchetan Dontha
    Nov 24 at 15:27











    0














    An alternative approach: $frac{p}{p-1}geq 1+frac{1}{p}$ gives
    $$ prod_{p}frac{p}{p-1}geq prod_{p}left(1+frac{1}{p}right) =!!!!!!!! sum_{substack{ngeq 1\ntext{ squarefree}}}!!!!!!!frac{1}{n} $$
    but since the set of squarefree numbers has a positive density in $mathbb{N}$ ($frac{6}{pi^2}$, which is also the probability that two "random integers" are coprime), by summation by parts it follows that
    $$sum_{substack{1leq nleq N\ntext{ squarefree}}}!!!!frac{1}{n}=O(1)+frac{6}{pi^2}sum_{n=1}^{N}frac{1}{n}=frac{6}{pi^2}log(N)+O(1) $$
    so the original product is divergent.






    share|cite|improve this answer


























      0














      An alternative approach: $frac{p}{p-1}geq 1+frac{1}{p}$ gives
      $$ prod_{p}frac{p}{p-1}geq prod_{p}left(1+frac{1}{p}right) =!!!!!!!! sum_{substack{ngeq 1\ntext{ squarefree}}}!!!!!!!frac{1}{n} $$
      but since the set of squarefree numbers has a positive density in $mathbb{N}$ ($frac{6}{pi^2}$, which is also the probability that two "random integers" are coprime), by summation by parts it follows that
      $$sum_{substack{1leq nleq N\ntext{ squarefree}}}!!!!frac{1}{n}=O(1)+frac{6}{pi^2}sum_{n=1}^{N}frac{1}{n}=frac{6}{pi^2}log(N)+O(1) $$
      so the original product is divergent.






      share|cite|improve this answer
























        0












        0








        0






        An alternative approach: $frac{p}{p-1}geq 1+frac{1}{p}$ gives
        $$ prod_{p}frac{p}{p-1}geq prod_{p}left(1+frac{1}{p}right) =!!!!!!!! sum_{substack{ngeq 1\ntext{ squarefree}}}!!!!!!!frac{1}{n} $$
        but since the set of squarefree numbers has a positive density in $mathbb{N}$ ($frac{6}{pi^2}$, which is also the probability that two "random integers" are coprime), by summation by parts it follows that
        $$sum_{substack{1leq nleq N\ntext{ squarefree}}}!!!!frac{1}{n}=O(1)+frac{6}{pi^2}sum_{n=1}^{N}frac{1}{n}=frac{6}{pi^2}log(N)+O(1) $$
        so the original product is divergent.






        share|cite|improve this answer












        An alternative approach: $frac{p}{p-1}geq 1+frac{1}{p}$ gives
        $$ prod_{p}frac{p}{p-1}geq prod_{p}left(1+frac{1}{p}right) =!!!!!!!! sum_{substack{ngeq 1\ntext{ squarefree}}}!!!!!!!frac{1}{n} $$
        but since the set of squarefree numbers has a positive density in $mathbb{N}$ ($frac{6}{pi^2}$, which is also the probability that two "random integers" are coprime), by summation by parts it follows that
        $$sum_{substack{1leq nleq N\ntext{ squarefree}}}!!!!frac{1}{n}=O(1)+frac{6}{pi^2}sum_{n=1}^{N}frac{1}{n}=frac{6}{pi^2}log(N)+O(1) $$
        so the original product is divergent.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 24 at 18:24









        Jack D'Aurizio

        285k33277655




        285k33277655






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3011661%2fsolution-to-infinite-product-prod-p-primes-infty-fracpp-1%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Aardman Animations

            Are they similar matrix

            “minimization” problem in Euclidean space related to orthonormal basis