Showing that $prod_{k=1} ^{infty}left(1- frac{1}{2k}right) = 0$ [closed]












4














How to find this product?



$$prod_{k=1} ^{infty}left(1- frac{1}{2k}right)$$



I know the answer is $0$, but I want to know how and why.






sequences-and-series products infinite-product







share|cite|improve this question















closed as off-topic by Saad, Jean-Claude Arbaut, Vidyanshu Mishra, José Carlos Santos, Omran Kouba Dec 1 at 14:21


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, Jean-Claude Arbaut, Vidyanshu Mishra, José Carlos Santos, Omran Kouba

If this question can be reworded to fit the rules in the help center, please edit the question.













  • This may help: math.stackexchange.com/questions/788096/…
    – Jean-Claude Arbaut
    Dec 1 at 10:49












  • $$frac{(2n-1)!!}{(2n)!!}=frac{1}{4^n}binom{2n}{n}leq frac{1}{sqrt{pi n}}.$$
    – Jack D'Aurizio
    Dec 1 at 11:36












  • That’s really curious that OP was closed because off topic and the one related indicated by JC Arbaut is considered a perfect OP. I think that the revisors should reconsider their guide lines to make such kind of distinctions!
    – gimusi
    Dec 1 at 14:24
















4














How to find this product?



$$prod_{k=1} ^{infty}left(1- frac{1}{2k}right)$$



I know the answer is $0$, but I want to know how and why.






sequences-and-series products infinite-product







share|cite|improve this question















closed as off-topic by Saad, Jean-Claude Arbaut, Vidyanshu Mishra, José Carlos Santos, Omran Kouba Dec 1 at 14:21


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, Jean-Claude Arbaut, Vidyanshu Mishra, José Carlos Santos, Omran Kouba

If this question can be reworded to fit the rules in the help center, please edit the question.













  • This may help: math.stackexchange.com/questions/788096/…
    – Jean-Claude Arbaut
    Dec 1 at 10:49












  • $$frac{(2n-1)!!}{(2n)!!}=frac{1}{4^n}binom{2n}{n}leq frac{1}{sqrt{pi n}}.$$
    – Jack D'Aurizio
    Dec 1 at 11:36












  • That’s really curious that OP was closed because off topic and the one related indicated by JC Arbaut is considered a perfect OP. I think that the revisors should reconsider their guide lines to make such kind of distinctions!
    – gimusi
    Dec 1 at 14:24














4












4








4


1





How to find this product?



$$prod_{k=1} ^{infty}left(1- frac{1}{2k}right)$$



I know the answer is $0$, but I want to know how and why.






sequences-and-series products infinite-product







share|cite|improve this question















How to find this product?



$$prod_{k=1} ^{infty}left(1- frac{1}{2k}right)$$



I know the answer is $0$, but I want to know how and why.






sequences-and-series products infinite-product




sequences-and-series products infinite-product






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 at 15:23









Martin Sleziak

44.6k7115270




44.6k7115270










asked Dec 1 at 10:02









S. Yoo

453




453




closed as off-topic by Saad, Jean-Claude Arbaut, Vidyanshu Mishra, José Carlos Santos, Omran Kouba Dec 1 at 14:21


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, Jean-Claude Arbaut, Vidyanshu Mishra, José Carlos Santos, Omran Kouba

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Saad, Jean-Claude Arbaut, Vidyanshu Mishra, José Carlos Santos, Omran Kouba Dec 1 at 14:21


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, Jean-Claude Arbaut, Vidyanshu Mishra, José Carlos Santos, Omran Kouba

If this question can be reworded to fit the rules in the help center, please edit the question.












  • This may help: math.stackexchange.com/questions/788096/…
    – Jean-Claude Arbaut
    Dec 1 at 10:49












  • $$frac{(2n-1)!!}{(2n)!!}=frac{1}{4^n}binom{2n}{n}leq frac{1}{sqrt{pi n}}.$$
    – Jack D'Aurizio
    Dec 1 at 11:36












  • That’s really curious that OP was closed because off topic and the one related indicated by JC Arbaut is considered a perfect OP. I think that the revisors should reconsider their guide lines to make such kind of distinctions!
    – gimusi
    Dec 1 at 14:24


















  • This may help: math.stackexchange.com/questions/788096/…
    – Jean-Claude Arbaut
    Dec 1 at 10:49












  • $$frac{(2n-1)!!}{(2n)!!}=frac{1}{4^n}binom{2n}{n}leq frac{1}{sqrt{pi n}}.$$
    – Jack D'Aurizio
    Dec 1 at 11:36












  • That’s really curious that OP was closed because off topic and the one related indicated by JC Arbaut is considered a perfect OP. I think that the revisors should reconsider their guide lines to make such kind of distinctions!
    – gimusi
    Dec 1 at 14:24
















This may help: math.stackexchange.com/questions/788096/…
– Jean-Claude Arbaut
Dec 1 at 10:49






This may help: math.stackexchange.com/questions/788096/…
– Jean-Claude Arbaut
Dec 1 at 10:49














$$frac{(2n-1)!!}{(2n)!!}=frac{1}{4^n}binom{2n}{n}leq frac{1}{sqrt{pi n}}.$$
– Jack D'Aurizio
Dec 1 at 11:36






$$frac{(2n-1)!!}{(2n)!!}=frac{1}{4^n}binom{2n}{n}leq frac{1}{sqrt{pi n}}.$$
– Jack D'Aurizio
Dec 1 at 11:36














That’s really curious that OP was closed because off topic and the one related indicated by JC Arbaut is considered a perfect OP. I think that the revisors should reconsider their guide lines to make such kind of distinctions!
– gimusi
Dec 1 at 14:24




That’s really curious that OP was closed because off topic and the one related indicated by JC Arbaut is considered a perfect OP. I think that the revisors should reconsider their guide lines to make such kind of distinctions!
– gimusi
Dec 1 at 14:24










2 Answers
2






active

oldest

votes


















6














We have that by $log (1-x)le -x$



$$logprod_{k=1} ^{n}{left(1- frac{1}{2k}right)}=sum_{k=1} ^{n} logleft(1- frac{1}{2k}right)le -sum_{k=1} ^{n} frac1{2k}to -infty$$






share|cite|improve this answer































    2














    Note that



    $$begin{align}prod_{k=1}^inftyleft(1-frac1{2k}right)&=lim_{ntoinfty}prod_{k=1}^nleft(frac{2k-1}{2k}right)\
    &=lim_{ntoinfty}frac{1cdot 3cdots(2n-1)}{2cdot 4cdots 2n}\
    &=lim_{ntoinfty}frac{1cdot 3cdots(2n-1)}{2cdot 4cdots 2n}cdotfrac{2cdot 4cdots 2n}{2cdot 4cdots 2n}\
    &=lim_{ntoinfty}frac{(2n)!}{(2^n(1cdot 2cdots n))^2}\
    &=lim_{ntoinfty}frac{(2n)!}{4^n(n!)^2}end{align}$$



    Now using the Stirling approximation in the last expression completes the proof.






    share|cite|improve this answer



















    • 1




      Actually the $"expsumlog"$ stuff is useless here, as the final formula comes from the product. But +1 anyway. And even without Stirling, one cas use Wallis' integrals (actually a way to prove Stirling's formula).
      – Jean-Claude Arbaut
      Dec 1 at 10:45












    • ah, you are totally right @Jean-Claude, I wrote this senseless
      – Masacroso
      Dec 1 at 10:56


















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6














    We have that by $log (1-x)le -x$



    $$logprod_{k=1} ^{n}{left(1- frac{1}{2k}right)}=sum_{k=1} ^{n} logleft(1- frac{1}{2k}right)le -sum_{k=1} ^{n} frac1{2k}to -infty$$






    share|cite|improve this answer




























      6














      We have that by $log (1-x)le -x$



      $$logprod_{k=1} ^{n}{left(1- frac{1}{2k}right)}=sum_{k=1} ^{n} logleft(1- frac{1}{2k}right)le -sum_{k=1} ^{n} frac1{2k}to -infty$$






      share|cite|improve this answer


























        6












        6








        6






        We have that by $log (1-x)le -x$



        $$logprod_{k=1} ^{n}{left(1- frac{1}{2k}right)}=sum_{k=1} ^{n} logleft(1- frac{1}{2k}right)le -sum_{k=1} ^{n} frac1{2k}to -infty$$






        share|cite|improve this answer














        We have that by $log (1-x)le -x$



        $$logprod_{k=1} ^{n}{left(1- frac{1}{2k}right)}=sum_{k=1} ^{n} logleft(1- frac{1}{2k}right)le -sum_{k=1} ^{n} frac1{2k}to -infty$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 1 at 10:21

























        answered Dec 1 at 10:11









        gimusi

        1




        1























            2














            Note that



            $$begin{align}prod_{k=1}^inftyleft(1-frac1{2k}right)&=lim_{ntoinfty}prod_{k=1}^nleft(frac{2k-1}{2k}right)\
            &=lim_{ntoinfty}frac{1cdot 3cdots(2n-1)}{2cdot 4cdots 2n}\
            &=lim_{ntoinfty}frac{1cdot 3cdots(2n-1)}{2cdot 4cdots 2n}cdotfrac{2cdot 4cdots 2n}{2cdot 4cdots 2n}\
            &=lim_{ntoinfty}frac{(2n)!}{(2^n(1cdot 2cdots n))^2}\
            &=lim_{ntoinfty}frac{(2n)!}{4^n(n!)^2}end{align}$$



            Now using the Stirling approximation in the last expression completes the proof.






            share|cite|improve this answer



















            • 1




              Actually the $"expsumlog"$ stuff is useless here, as the final formula comes from the product. But +1 anyway. And even without Stirling, one cas use Wallis' integrals (actually a way to prove Stirling's formula).
              – Jean-Claude Arbaut
              Dec 1 at 10:45












            • ah, you are totally right @Jean-Claude, I wrote this senseless
              – Masacroso
              Dec 1 at 10:56
















            2














            Note that



            $$begin{align}prod_{k=1}^inftyleft(1-frac1{2k}right)&=lim_{ntoinfty}prod_{k=1}^nleft(frac{2k-1}{2k}right)\
            &=lim_{ntoinfty}frac{1cdot 3cdots(2n-1)}{2cdot 4cdots 2n}\
            &=lim_{ntoinfty}frac{1cdot 3cdots(2n-1)}{2cdot 4cdots 2n}cdotfrac{2cdot 4cdots 2n}{2cdot 4cdots 2n}\
            &=lim_{ntoinfty}frac{(2n)!}{(2^n(1cdot 2cdots n))^2}\
            &=lim_{ntoinfty}frac{(2n)!}{4^n(n!)^2}end{align}$$



            Now using the Stirling approximation in the last expression completes the proof.






            share|cite|improve this answer



















            • 1




              Actually the $"expsumlog"$ stuff is useless here, as the final formula comes from the product. But +1 anyway. And even without Stirling, one cas use Wallis' integrals (actually a way to prove Stirling's formula).
              – Jean-Claude Arbaut
              Dec 1 at 10:45












            • ah, you are totally right @Jean-Claude, I wrote this senseless
              – Masacroso
              Dec 1 at 10:56














            2












            2








            2






            Note that



            $$begin{align}prod_{k=1}^inftyleft(1-frac1{2k}right)&=lim_{ntoinfty}prod_{k=1}^nleft(frac{2k-1}{2k}right)\
            &=lim_{ntoinfty}frac{1cdot 3cdots(2n-1)}{2cdot 4cdots 2n}\
            &=lim_{ntoinfty}frac{1cdot 3cdots(2n-1)}{2cdot 4cdots 2n}cdotfrac{2cdot 4cdots 2n}{2cdot 4cdots 2n}\
            &=lim_{ntoinfty}frac{(2n)!}{(2^n(1cdot 2cdots n))^2}\
            &=lim_{ntoinfty}frac{(2n)!}{4^n(n!)^2}end{align}$$



            Now using the Stirling approximation in the last expression completes the proof.






            share|cite|improve this answer














            Note that



            $$begin{align}prod_{k=1}^inftyleft(1-frac1{2k}right)&=lim_{ntoinfty}prod_{k=1}^nleft(frac{2k-1}{2k}right)\
            &=lim_{ntoinfty}frac{1cdot 3cdots(2n-1)}{2cdot 4cdots 2n}\
            &=lim_{ntoinfty}frac{1cdot 3cdots(2n-1)}{2cdot 4cdots 2n}cdotfrac{2cdot 4cdots 2n}{2cdot 4cdots 2n}\
            &=lim_{ntoinfty}frac{(2n)!}{(2^n(1cdot 2cdots n))^2}\
            &=lim_{ntoinfty}frac{(2n)!}{4^n(n!)^2}end{align}$$



            Now using the Stirling approximation in the last expression completes the proof.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 1 at 21:41

























            answered Dec 1 at 10:09









            Masacroso

            12.8k41746




            12.8k41746








            • 1




              Actually the $"expsumlog"$ stuff is useless here, as the final formula comes from the product. But +1 anyway. And even without Stirling, one cas use Wallis' integrals (actually a way to prove Stirling's formula).
              – Jean-Claude Arbaut
              Dec 1 at 10:45












            • ah, you are totally right @Jean-Claude, I wrote this senseless
              – Masacroso
              Dec 1 at 10:56














            • 1




              Actually the $"expsumlog"$ stuff is useless here, as the final formula comes from the product. But +1 anyway. And even without Stirling, one cas use Wallis' integrals (actually a way to prove Stirling's formula).
              – Jean-Claude Arbaut
              Dec 1 at 10:45












            • ah, you are totally right @Jean-Claude, I wrote this senseless
              – Masacroso
              Dec 1 at 10:56








            1




            1




            Actually the $"expsumlog"$ stuff is useless here, as the final formula comes from the product. But +1 anyway. And even without Stirling, one cas use Wallis' integrals (actually a way to prove Stirling's formula).
            – Jean-Claude Arbaut
            Dec 1 at 10:45






            Actually the $"expsumlog"$ stuff is useless here, as the final formula comes from the product. But +1 anyway. And even without Stirling, one cas use Wallis' integrals (actually a way to prove Stirling's formula).
            – Jean-Claude Arbaut
            Dec 1 at 10:45














            ah, you are totally right @Jean-Claude, I wrote this senseless
            – Masacroso
            Dec 1 at 10:56




            ah, you are totally right @Jean-Claude, I wrote this senseless
            – Masacroso
            Dec 1 at 10:56



            -

            Popular posts from this blog

            Berounka

            Image
            Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc...
            Read more

            Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

            Image
            3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
            Read more

            Sphinx de Gizeh

            Image
            Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
            Read more